Notebook

Convolution Integral¶

A convolution, $(h*x)(t)$ , is a mathematical operator which manipulates two functions $x$ and $h$ and produces an output that represents the amount of overlap between $x$ and a reversed and translated version of $h$ .

In signal processing, one of the functions ( $h$ ) is taken to be a fixed filter impulse response, and is known as a kernel.¶

$(x \ast h)(t) = \int_{-\infty}^{\infty} x(\tau)h(t - \tau) \, d\tau$

$(h \ast x)(t) = \int_{-\infty}^{\infty} h(\tau)x(t - \tau) \, d\tau$

Illustration: [1]

In [188]:

from IPython.display import HTML
HTML('<iframe src=http://mathworld.wolfram.com/images/gifs/convrect.gif width=300 height=260></iframe>')

Out[188]:

In [189]:

from IPython.display import HTML
HTML('<iframe src=http://mathworld.wolfram.com/images/gifs/convgaus.gif width=300 height=260></iframe>')

Out[189]:

Steps

1. The impulse response $h(\tau)$ is flipped or time-reversed (that is, reflected about the origin) to obtain $h(-\tau)$ and then shifted by to form $h(t - \tau) = h[-(\tau - t)]$ , which is a function of $\tau$ with parameter $t$ .¶

2. The signal $x(\tau)$ and $h(t - \tau)$ are multiplied together for all values of $\tau$ with $t$ fixed at some value.¶

3. The product $x(\tau)h(t - \tau)$ is integrated over all $\tau$ to produce a single output value $y(t)$ .¶

Ex.

In [190]:

def step(t, t_0):
    return 1 * ((t - t_0) >= 0)

In [191]:

%matplotlib inline  
import matplotlib.pyplot as plt
import numpy as np

Basic Example:¶

Calculate the convolution of a unit step input with a unit step response system.¶

In [192]:

t = np.arange(-4.0, 4.0, 0.01)
x = step(t,0) 
h = step(t,0)

fig = plt.figure(figsize=(12,4))

plt.subplot(1, 2, 1)
plt.plot(t,x,'g-',linewidth=6);
plt.title('Input signal, $x(t)$')
plt.xlabel('t')

plt.subplot(1, 2, 2)
plt.plot(t,h,'r-',linewidth=6);
plt.title('Impulse response, $h(t)$')
plt.xlabel('t');

$(x \ast h)(t) = \int_{-\infty}^{\infty} u(\tau)u(t - \tau) \, d\tau$

Step 1: The impulse response $h(\tau)$ is flipped or time-reversed (that is, reflected about the origin) to obtain $h(-\tau)$ and then shifted by to form $h(t - \tau) = h[-(\tau - t)]$ , which is a function of $\tau$ with parameter $t$ .¶

In [193]:

t = np.arange(-4.0, 4.0, 0.01)
h = step(-t, 0)
fig = plt.figure(figsize=(12,4))
plt.plot(t,h,'r-',linewidth=6);
plt.title('Impulse response, $h(-\u03c4)$ at t = 0')
plt.xlabel('T');

Step 2-3 The signal $x(\tau)$ and $h(t - \tau)$ are multiplied together for all values of $\tau$ with $t$ fixed at some value. Then, the product $x(\tau)h(t - \tau)$ is integrated over all $\tau$ to produce a single output value $y(t)$ .¶

In [194]:

t = np.arange(-4.0, 4.0, 0.01)
h = step(-t, -0.5)
x = step(t, 0)
fig = plt.figure(figsize=(12,4))
ax1 = fig.add_axes([0, 0, 1, 1])   # [left, bottom, width, height]

plt.plot(t,h,'r-',linewidth=6);
plt.plot(t,x,'g-',linewidth=6);
plt.title('Impulse response, $h(-\u03c4)$ at t = 0.5 with the input signal, $u(t)$')
plt.xlabel('T');
ax1.fill_between(t,0,h,where=x==h,color='black');
ax1.annotate('$t$', xy=(0.5, 0), xytext=(0.6, 0.1),
            arrowprops=dict(facecolor='yellow', shrink=0.05),
            )

Out[194]:

<matplotlib.text.Annotation at 0xb00f372c>

Notice that the time reversed Impulse response, $u(-\tau)$ , starts intersecting the signal, which is also a unit step, $u(\tau)$ , at $\tau = 0$ until $\tau = t$ .

Thus,

$(x \ast h)(t) = \int_{0}^{t} x(\tau)h(t - \tau) \, d\tau$

$(x \ast h)(t) = \int_{0}^{t} u(\tau)u(t - \tau) \, d\tau$

$(x \ast h)(t) = \int_{0}^{t} (1)(1) \, d\tau$

$(x \ast h)(t) = \left. \tau \right |_0^t$

$(x \ast h)(t) = t \, u(t)$

$(x \ast h)(t) = \begin{cases} 0 & \mbox{if } -\infty < t < 0 \\ t & \mbox{if } \, 0 < t < \infty \\ \end{cases}$

In [195]:

t = np.arange(-4.0, 4.0, 0.01)

# input
x = step(t,0)  

# impulse response
h = step(t,0) 

# result
result = np.zeros(len(t))
result = t * (step(t,0))

fig = plt.figure(figsize=(16,4))

plt.plot(t,x,'go-',linewidth=5, label='Input signal, $x(t)$');
plt.plot(t,h,'rd-',linewidth=1, label='Impulse response, $h(t)$');
plt.plot(t,result,'b-',linewidth=3, label='Output signal, $y(t)$');
plt.xlabel('t')
plt.grid()
legend = plt.legend(loc='upper left');

In [196]:

t = np.arange(-4.0, 4.0, 0.01)
x = step(t,1) - step(t,3)
h = (t + 2)*(step(t,-2) - step(t,-1))

fig = plt.figure(figsize=(12,4))

plt.subplot(1, 2, 1)
plt.plot(t,x,'g-',linewidth=5);
plt.title('Input signal, $x(t)$')
plt.xlabel('t')

plt.subplot(1, 2, 2)
plt.plot(t,h,'r-',linewidth=3);
plt.title('Impulse response, $h(t)$')
plt.xlabel('t');

Drill Problem 1: Compute the convolution, $(x \ast h)(t)$ , of the signals represented in the figure above:¶

Step 1: The impulse response $h(\tau)$ is flipped or time-reversed (that is, reflected about the origin) to obtain $h(-\tau)$ and then shifted by to form $h(t - \tau) = h[-(\tau - t)]$ , which is a function of $\tau$ with parameter $t$ .¶

In [197]:

t = np.arange(-4.0, 4.0, 0.01)
h = (-t + 2)*(step(t,1) - step(t,2))
fig = plt.figure(figsize=(12,4))
plt.plot(t,h,'r-',linewidth=3);
plt.title('Impulse response, $h(-\u03c4)$ at t = 0')
plt.xlabel('T');

Note: This is the $h(t - \tau)$ at $t = 0$ and the signal starts at $\tau = t + 1$ and ends at $\tau = t + 2$ .

Step 2-3 The signal $x(\tau)$ and $h(t - \tau)$ are multiplied together for all values of $\tau$ with $t$ fixed at some value. Then, the product $x(\tau)h(t - \tau)$ is integrated over all $\tau$ to produce a single output value $y(t)$ .¶

Case I: NO Overlap, $y(t) = 0$ at $-\infty \to t_1$ , where $t_1 = -1$ ¶

$y(t) = \int_{-\infty}^{-1} x(\tau)h(t - \tau) \, d\tau = 0$

In [198]:

t = np.arange(-4.0, 4.0, 0.01)
h_4 = (-t - 2)*(step(t,-3) - step(t,-2))
h_3 = (-t - 1)*(step(t,-2) - step(t,-1))
h_2 = (-t - 0)*(step(t,-1) - step(t,-0))
h_1 = (-t + 1)*(step(t,0) - step(t,1))
fig = plt.figure(figsize=(16,4))
plt.plot(t,h_4,'r-',linewidth=3, label="$t=-4$");
plt.plot(t,h_3,'c-',linewidth=3, label="$t=-3$");
plt.plot(t,h_2,'m-',linewidth=3, label="$t=-2$");
plt.plot(t,h_1,'y-',linewidth=3, label="$t=-1$");
plt.title('Impulse response, $h(-(\u03c4 - t))$ at t = -4,-3,-2,-1, with the input signal, $x(t)$')
plt.plot(t,x,'g-',linewidth=5, label="$x(t)$");
plt.xlabel('T');

legend = plt.legend() 

Case II: Leading-edge overlap, at $t_1 \to t_2$ or at $-1 < t < 0$ ¶

$y(t) = \int_{a}^{b} x(\tau)h(t - \tau) \, d\tau$

In [199]:

t = np.arange(-4.0, 4.0, 0.01)
h_half = (-t + 1.5)*(step(t,0.5) - step(t,1.5))
fig = plt.figure(figsize=(12,2.5))
ax1 = fig.add_axes([0, 0, 1, 1])   # [left, bottom, width, height]
plt.plot(t,h_half,'r-',linewidth=3, label="$t=0.5$");
plt.title('Impulse response, $h(-(\u03c4 - t))$ at t = -0.5, with the input signal, $x(t)$')
plt.plot(t,x,'g-',linewidth=5, label="$x(t)$");
plt.xlabel('T');
ax1.fill_between(t,0,h_half,where=x>=h_half,color='black')
ax1.annotate('$t + 2$', xy=(1.5, 0), xytext=(1.6, 0.1),
            arrowprops=dict(facecolor='yellow', shrink=0.05),
            )
ax1.annotate('same \n length \n as \n base', xy=(1, 0.2), xytext=(0.55, 0.2),
            arrowprops=dict(facecolor='yellow'),
            )
ax1.annotate('t+2-1', xy=(1.25, 0), xytext=(1.1, -0.15),
            arrowprops=dict(facecolor='yellow', shrink=0.05),
            )
legend = plt.legend() 

The overlap starts at $\tau = 1$ and ends at $\tau = t + 2$ . Recall that the impulse response ends at $t+2$ . Thus,

$y(t) = \int_{1}^{t+2} x(\tau)h(t - \tau) \, d\tau$

In the above interval, $x(\tau) = 1$ and $h(t - \tau) = - \tau + t + 2$ .

$y(t) = \int_{1}^{t+2} (1)(- \tau + t + 2) \, d\tau$

$y(t) = \left. (t + 2) \tau - \dfrac{\tau^2}{2} \right |_1^{t+2}$

$y(t) = \left((t + 2)(t + 2) - \dfrac{(t+2)^2}{2}\right) - \left((t + 2) - \dfrac{1^2}{2}\right)$

$y(t) = \dfrac{(t+2)^2}{2} - t - \dfrac{3}{2}$

$y(t) = \dfrac{t^2}{2} + \dfrac{4t}{2} + \dfrac{4}{2} -t - \dfrac{3}{2}$

$y(t) = \dfrac{t^2}{2} + t + \dfrac{1}{2}$

$y(t) = \dfrac{1}{2} (t + 1)^2$

Alternatively, this can be evaluated by the triangle area:

$A = \dfrac{1}{2} bh$

$A = \dfrac{1}{2} (t+2-1)(t+1)$

$A = \dfrac{1}{2}(t+1)^2$

Case III: Full overlap, at $t = 0$ to $t = 1$ ¶

$y(t) = \int_{a}^{b} x(\tau)h(t - \tau) \, d\tau$

In [200]:

t = np.arange(-4.0, 4.0, 0.01)
h_half = (-t + 2.5)*(step(t,1.5) - step(t,2.5))
fig = plt.figure(figsize=(12,2.5))
ax1 = fig.add_axes([0, 0, 1, 1])   # [left, bottom, width, height]
plt.plot(t,h_half,'r-',linewidth=3, label="$h$ at $t=0.5$");
plt.title('Impulse response, $h(-(\u03c4 - t))$ at t = -0.5, with the input signal, $x(t)$')
plt.plot(t,x,'g-',linewidth=5, label="$x(t)$");
plt.xlabel('T');
ax1.fill_between(t,0,h_half,where=x>=h_half,color='black')
legend = plt.legend(loc='center') 
ax1.annotate('$t + 2$', xy=(2.5, 0), xytext=(2.6, 0.1),
            arrowprops=dict(facecolor='yellow', shrink=0.05),
            )
ax1.annotate('$t + 1$', xy=(1.5, 0), xytext=(1.2, 0.1),
            arrowprops=dict(facecolor='yellow', shrink=0.05),
            );

The overlap starts at $\tau = t + 1$ and ends at $\tau = t + 2$ .

$y(t) = \int_{t+1}^{t+2} x(\tau)h(t - \tau) \, d\tau$

In the above interval, $x(\tau) = 1$ and $h(t - \tau) = - \tau + t + 2$ . Thus,

$y(t) = \int_{t+1}^{t+2} (1) (- \tau + t + 2) \, d\tau$

$y(t) = \left. (t + 2) \tau - \dfrac{\tau^2}{2} \right |_{t+1}^{t+2}$

$y(t) = \left((t + 2)(t + 2) - \dfrac{(t+2)^2}{2}\right) - \left((t + 2)(t+1) - \dfrac{(t+1)^2}{2}\right)$

$y(t) = \dfrac{(t+2)^2}{2} - (t + 1)(t+2 - \dfrac{t+1}{2})$

$y(t) = \dfrac{(t^2 + 4t + 4)}{2} - (t + 1)(\dfrac{t}{2} + \dfrac{3}{2})$

$y(t) = \dfrac{t^2}{2} + \dfrac{4t}{2} + \dfrac{4}{2} - \dfrac{t^2}{2} - \dfrac{4t}{2} - \dfrac{3}{2}$

$y(t) = \dfrac{1}{2}$

Alternatively, this can be evaluated by the triangle area:

$A = \dfrac{1}{2} bh$

$A = \dfrac{1}{2} (1)(1)$

$A = \dfrac{1}{2}$

Case IV: Trailing edge overlap, at $1 < t < 2$ ¶

$y(t) = \int_{a}^{b} x(\tau)h(t - \tau) \, d\tau$

In [201]:

t = np.arange(-4.0, 4.0, 0.01)
h_half = (-t + 3.5)*(step(t,2.5) - step(t,3.5))
fig = plt.figure(figsize=(12,2.5))
ax1 = fig.add_axes([0, 0, 1, 1])   # [left, bottom, width, height]
plt.plot(t,h_half,'r-',linewidth=3, label="$h$ at $t=0.5$");
plt.title('Impulse response, $h(-(\u03c4 - t))$ at t = -0.5, with the input signal, $x(t)$')
plt.plot(t,x,'g-',linewidth=5, label="$x(t)$");
plt.xlabel('T');
ax1.fill_between(t,0,h_half,where=x>=h_half,color='black')
legend = plt.legend(loc='center') 
ax1.annotate('$t + 1$', xy=(2.5, 0), xytext=(2.2, 0.1),
            arrowprops=dict(facecolor='yellow', shrink=0.05),
            );

The overlap starts at $\tau = t + 1$ and ends at $\tau = 3$ .

$y(t) = \int_{t+1}^{3} x(\tau)h(t - \tau) \, d\tau$

In the above interval, $x(\tau) = 1$ and $h(t - \tau) = - \tau + t + 2$ . Thus,

$y(t) = \int_{t+1}^{3} (1) (- \tau + t + 2) \, d\tau$

$y(t) = \left. (t + 2) \tau - \dfrac{\tau^2}{2} \right |_{t+1}^{3}$

$y(t) = \left((t + 2)(3) - \dfrac{(3)^2}{2}\right) - \left((t + 2)(t + 1) - \dfrac{(t+1)^2}{2}\right)$

$y(t) = \left(3t + 6 - \dfrac{9}{2}\right) - \left((t + 1)((t+2)- \dfrac{(t+1)}{2} )\right)$

$y(t) = (3t + \dfrac{3}{2}) - (t + 1)(\dfrac{t}{2} + \dfrac{3}{2})$

$y(t) = 3t + \dfrac{3}{2} - \dfrac{t^2}{2} - \dfrac{4t}{2} - \dfrac{3}{2}$

$y(t) = t - \dfrac{t^2}{2}$

$y(t) = t (1 - \dfrac{t}{2})$

Alternatively, this can be evaluated by subtracting the smaller triangle area, $A_t$ , from the bigger triangle, $A_T$ :

$A = A_T - A_t$

$A = \dfrac{1}{2} - \dfrac{1}{2} bh$

$A = \dfrac{1}{2} - \dfrac{1}{2} (t+2-3)(t+2-3)$

$A = \dfrac{1}{2} - \dfrac{1}{2}(t-1)^2$

$A = t - \dfrac{t^2}{2}$

Case V: No overlap, at $t > 2$ ¶

$y(t) = \int_{2}^{\infty} x(\tau)h(t - \tau) \, d\tau = 0$

In [202]:

t = np.arange(-4.0, 4.0, 0.01)
h_half = (-t + 4)*(step(t,3) - step(t,4))
fig = plt.figure(figsize=(12,2.5))
ax1 = fig.add_axes([0, 0, 1, 1])   # [left, bottom, width, height]
plt.plot(t,h_half,'r-',linewidth=3, label="$h$ at $t=2$");
plt.title('Impulse response, $h(-(\u03c4 - t))$ at t = -0.5, with the input signal, $x(t)$')
plt.plot(t,x,'g-',linewidth=5, label="$x(t)$");
plt.xlabel('T');
ax1.fill_between(t,0,h_half,where=x>=h_half,color='black')
legend = plt.legend(loc='center') 
ax1.annotate('No Overlap', xy=(3, 0.4), xytext=(2.2, 0.4),
            arrowprops=dict(facecolor='yellow', shrink=0.05),
            );

1. Final Answer¶

Therefore, the convolution result is:

$y(t) = \begin{cases} 0 & \mbox{if } -\infty < t < -1 \\ \frac{1}{2}(t+1)^2 & \mbox{if } -1 < t < 0 \\ \frac{1}{2} & \mbox{if } \, 0 < t < 1 \\ t - \frac{t^2}{2} & \mbox{if } \, 1 < t < 2 \\ 0 & \mbox{if } \, 2 < t < \infty \\ \end{cases}$

In [203]:

t = np.arange(-4.0, 4.0, 0.01)

# input
x = step(t,1) - step(t,3)

# impulse response
h = (t + 2)*(step(t,-2) - step(t,-1))

# result
result = np.zeros(len(t))
result = 0.5*(t+1)**2 * (step(t,-1)-step(t,0)) + 0.5* (step(t,0)-step(t,1)) + t * (1 - 0.5*t) * (step(t,1)-step(t,2))

fig = plt.figure(figsize=(14,4))

plt.plot(t,x,'g-',linewidth=5, label='Input signal, $x(t)$');
plt.plot(t,h,'r-',linewidth=3, label='Impulse response, $h(t)$');
plt.plot(t,result,'b-',linewidth=3, label='Output signal, $y(t)$');
plt.xlabel('t')
plt.grid()
legend = plt.legend(loc='upper left');

In [204]:

t = np.arange(-2.0, 2.0, 0.01)
x = step(t,0) - step(t,1)
h = (-t + 1)*(step(t,0) - step(t,1))

fig = plt.figure(figsize=(15,3))

plt.subplot(1, 2, 1)
plt.plot(t,x,'g-',linewidth=5);
plt.title('Input signal, $x(t)$')
plt.xlabel('t')

plt.subplot(1, 2, 2)
plt.plot(t,h,'r-',linewidth=3);
plt.title('Impulse response, $h(t)$')
plt.xlabel('t');

Drill Problem 2: Compute the convolution, $(h \ast x)(t)$ , of the signals represented in the figure above:¶

$y(t) = (h \ast x)(t) = \int_{-\infty}^{\infty} h(\tau)x(t - \tau) \, d\tau$

Note: In this exercise, the signal is time-reversed.

Step 1: The impulse response $h(\tau)$ is flipped or time-reversed (that is, reflected about the origin) to obtain $h(-\tau)$ and then shifted by to form $h(t - \tau) = h[-(\tau - t)]$ , which is a function of $\tau$ with parameter $t$ .¶

In [205]:

t = np.arange(-2.0, 2.0, 0.01)
x = step(t,-1) - step(t,0)

fig = plt.figure(figsize=(12,3))
ax1 = fig.add_axes([0, 0, 1, 1])   # [left, bottom, width, height]

plt.plot(t,x,'g-',linewidth=5);
plt.title('Time-reversed input signal, $x(t-\u03c4)$')
plt.xlabel('T');
ax1.annotate('$t$', xy=(0, 0), xytext=(0.1, 0.1),
            arrowprops=dict(facecolor='yellow', shrink=0.05),
            );
ax1.annotate('$t-1$', xy=(-1, 0), xytext=(-1.2, 0.1),
            arrowprops=dict(facecolor='yellow', shrink=0.05),
            );

Note: This rectangular pulse begins at $\tau = t-1$ and ends at $\tau = t$ .

Step 2-3 The signal $x(\tau)$ and $h(t - \tau)$ are multiplied together for all values of $\tau$ with $t$ fixed at some value. Then, the product $x(\tau)h(t - \tau)$ is integrated over all $\tau$ to produce a single output value $y(t)$ .¶

In [ ]:

In [206]:

t = np.arange(-2.0, 2.0, 0.01)
x_0 = step(t,-1) - step(t,0)
x__5 = step(t,-1.5) - step(t,-0.5)
x_1 = step(t,-2) - step(t,-1)
h = (-t + 1)*(step(t,0) - step(t,1))

fig = plt.figure(figsize=(12,3))
ax1 = fig.add_axes([1, 0, 1, 1])   # [left, bottom, width, height]

plt.plot(t,x_1,'cd-.',linewidth=5, label="x at $t=-1$");
plt.plot(t,x__5,'m--',linewidth=5, label="x at $t=-0.5$");
plt.plot(t,x_0,'g-',linewidth=3, label="x at $t=0$");
plt.title('Time-reversed input signal, $x(t-\tau)$ with the Impulse response, $h(t)$')
plt.plot(t,h,'r-',linewidth=3, label="h(t)");
plt.xlabel('T');

legend = plt.legend(loc='upper right');

Case I: NO Overlap, $y(t) = 0$ at $-\infty \to t_1$ , where $t_1 = 0$ ¶

$(h \ast x)(t) = \int_{-\infty}^{0} h(\tau)x(t - \tau) \, d\tau = 0$

Case II: Leading-edge overlap, at $t_1 \to t_2$ or at $0 < t < 1$ ¶

$y(t) = \int_{a}^{b} x(\tau)h(t - \tau) \, d\tau$

In [207]:

t = np.arange(-2.0, 2.0, 0.01)
x = step(t,-0.5) - step(t,0.5)
h = (-t + 1)*(step(t,0) - step(t,1))

fig = plt.figure(figsize=(12,3))
ax1 = fig.add_axes([1, 0, 1, 1])   # [left, bottom, width, height]

plt.plot(t,x,'g-',linewidth=5, label="x at $t=0.5$");
plt.title('Time-reversed input signal, $x(t-T)$ with the Impulse response, $h(t)$')
plt.plot(t,h,'r-',linewidth=3, label="h");
plt.xlabel('T');

ax1.annotate('$t$', xy=(0.5, 0), xytext=(0.6, 0.15),
            arrowprops=dict(facecolor='yellow', shrink=0.05),
            );

ax1.fill_between(t,0,h,where=x>=h,color='black')
legend = plt.legend(loc='upper right');

The overlap starts at $\tau = 0$ and ends at $\tau = t$ . Thus,

$y(t) = \int_{0}^{t} h(\tau)x(t - \tau) \, d\tau = 0$

In the above interval, $x(t -\tau) = 1$ and $h(\tau) = -\tau + 1$ .

$y(t) = \int_{0}^{t} (-\tau + 1) (1) \, d\tau$

$y(t) = \left. \dfrac{-\tau^2}{2} + \tau \right |_{0}^{t}$

$y(t) = \left( \dfrac{-t^2}{2} + t \right) - 0$

$y(t) = t - \dfrac{t^2}{2}$

$y(t) = t ( 1 - \dfrac{t}{2})$

Case III: Full overlap, at $t = 1$ ¶

In [208]:

t = np.arange(-2.0, 2.0, 0.01)
x = step(t,0) - step(t,1)
h = (-t + 1)*(step(t,0) - step(t,1))

fig = plt.figure(figsize=(12,3))
ax1 = fig.add_axes([1, 0, 1, 1])   # [left, bottom, width, height]

plt.plot(t,x,'g-',linewidth=5, label="x at $t=0.5$");
plt.title('Time-reversed input signal, $x(t-T)$ with the Impulse response, $h(t)$')
plt.plot(t,h,'r-',linewidth=3, label="h");
plt.xlabel('T');

ax1.annotate('$t$', xy=(1, 0), xytext=(1.1, 0.15),
            arrowprops=dict(facecolor='yellow', shrink=0.05),
            );

ax1.annotate('$t-1$', xy=(0, 0), xytext=(-0.2, 0.15),
            arrowprops=dict(facecolor='yellow', shrink=0.05),
            );

ax1.fill_between(t,0,h,where=x>=h,color='black')
legend = plt.legend(loc='upper right');

$y(t) = (h \ast x)(t) = \int_{t-1}^{t} h(\tau)x(t - \tau) \, d\tau$

$y(t) = \int_{t-1}^{t} (-\tau + 1) (1) \, d\tau$

$y(t) = \left. \dfrac{-\tau^2}{2} + \tau \right |_{t-1}^{t}$

$y(t) = \left( \dfrac{-t^2}{2} + t \right) - \left( \dfrac{-(t-1)^2}{2} + (t-1) \right)$

$y(t) = \dfrac{-t^2}{2} + t - (t-1)(\dfrac{-(t-1)}{2} + 1)$

$y(t) = \dfrac{-t^2}{2} + t - (t-1)(\dfrac{-t}{2} + \dfrac{1}{2} + 1)$

$y(t) = \dfrac{-t^2}{2} + t - ( \dfrac{-t^2}{2} + \dfrac{4t}{2} - \dfrac{3}{2} )$

$y(t) = \dfrac{3}{2} - t$

at $t = 1$ ,

$y(t) = (h \ast x)(t) = \int_{0}^{1} h(\tau)x(t - \tau) \, d\tau = \dfrac{3}{2} - 1 = \dfrac{1}{2}$

Case IV: Trailing edge overlap, at $1 < t < 2$ ¶

In [209]:

t = np.arange(-2.0, 2.0, 0.01)
x = step(t,0.5) - step(t,1.5)
h = (-t + 1)*(step(t,0) - step(t,1))

fig = plt.figure(figsize=(12,3))
ax1 = fig.add_axes([1, 0, 1, 1])   # [left, bottom, width, height]

plt.plot(t,x,'g-',linewidth=5, label="x at $t=1.5$");
plt.title('Time-reversed input signal, $x(t-T)$ with the Impulse response, $h(t)$')
plt.plot(t,h,'r-',linewidth=3, label="h");
plt.xlabel('T');

ax1.annotate('$t-1$', xy=(0.5, 0), xytext=(0.3, 0.15),
            arrowprops=dict(facecolor='yellow', shrink=0.05),
            );

ax1.fill_between(t,0,h,where=x>=h,color='black')
legend = plt.legend(loc='upper right');

$y(t) = (h \ast x)(t) = \int_{t-1}^{1} h(\tau)x(t - \tau) \, d\tau$

$y(t) = \int_{t-1}^{1} (-\tau + 1) (1) \, d\tau$

$y(t) = \left. \dfrac{-\tau^2}{2} + \tau \right |_{t-1}^{1}$

$y(t) = \left( \dfrac{-1^2}{2} +1 \right) - \left( \dfrac{-(t - 1)^2}{2} + (t - 1) \right)$

$y(t) = \dfrac{1}{2} - (t - 1)(\dfrac{-(t -1)}{2} + 1)$

$y(t) = \dfrac{1}{2} - (t - 1)(\dfrac{-t}{2} + \dfrac{1}{2} + 1)$

$y(t) = \dfrac{1}{2} - ( \dfrac{-t^2}{2} + \dfrac{4t}{2} - \dfrac{3}{2} )$

$y(t) = \dfrac{t^2}{2} - 2t + 2$

Case V: No overlap, at $t > 2$ ¶

In [210]:

t = np.arange(-2.0, 2.0, 0.01)
x = step(t,1) - step(t,2)
h = (-t + 1)*(step(t,0) - step(t,1))

fig = plt.figure(figsize=(12,3))
ax1 = fig.add_axes([1, 0, 1, 1])   # [left, bottom, width, height]

plt.plot(t,x,'g-',linewidth=3, label="x at $t=2$");
plt.title('Time-reversed input signal, $x(t-T)$ with the Impulse response, $h(t)$')
plt.plot(t,h,'r-',linewidth=3, label="h");
plt.xlabel('T');
legend = plt.legend(loc='upper right');

$y(t) = (h \ast x)(t) = \int_{2}^{\infty} h(\tau)x(t - \tau) \, d\tau = 0$

2. Final Answer¶

Therefore, the convolution result is:

$y(t) = \begin{cases} 0 & \mbox{if } -\infty < t < 0 \\ t ( 1 - \frac{t}{2}) & \mbox{if } 0 < t < 1 \\ \frac{1}{2} & \mbox{if } \, t = 1 \\ \frac{t^2}{2} - 2t + 2 & \mbox{if } \, 1 < t < 2 \\ 0 & \mbox{if } \, 2 < t < \infty \\ \end{cases}$

In [211]:

t = np.arange(-4.0, 4.0, 0.01)

# input
x = step(t,0) - step(t,1)

# impulse response
h = (-t + 1)*(step(t,0) - step(t,1))

# result
result = np.zeros(len(t))
result = t * (1 - 0.5*t) * (step(t,0) - step(t,1)) + (0.5*t*t - 2*t + 2)*(step(t,1) - step(t,2))

fig = plt.figure(figsize=(16,4))

plt.plot(t,x,'g-',linewidth=5, label='Input signal, $x(t)$');
plt.plot(t,h,'r-',linewidth=3, label='Impulse response, $h(t)$');
plt.plot(t,result,'b-',linewidth=3, label='Output signal, $y(t)$');
plt.xlabel('t')
plt.grid()
legend = plt.legend(loc='upper left');

By: Melvin Cabatuan melvincabatuan@gmail.com

This ipython notebook is licensed under the CC-BY-NC-SA license: http://creativecommons.org/licenses/by-nc-sa/4.0/

References:

[1] Weisstein, Eric W. "Convolution." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/Convolution.html 
[2] Notes from  Prof. Alan Oppenheim. http://ocw.mit.edu/resources/res-6-007-signals-and-systems-spring-2011/

Convolution Integral¶

A convolution, (h∗x)(t)(h*x)(t), is a mathematical operator which manipulates two functions xx and hh and produces an output that represents the amount of overlap between xx and a reversed and translated version of hh.

In signal processing, one of the functions (hh) is taken to be a fixed filter impulse response, and is known as a kernel.¶

Steps

1. The impulse response h(τ)h(\tau) is flipped or time-reversed (that is, reflected about the origin) to obtain h(−τ)h(-\tau) and then shifted by to form h(t−τ)=h[−(τ−t)]h(t - \tau) = h[-(\tau - t)], which is a function of τ\tau with parameter tt.¶

2. The signal x(τ)x(\tau) and h(t−τ)h(t - \tau) are multiplied together for all values of τ\tau with tt fixed at some value.¶

3. The product x(τ)h(t−τ)x(\tau)h(t - \tau) is integrated over all τ\tau to produce a single output value y(t)y(t).¶

Basic Example:¶

Calculate the convolution of a unit step input with a unit step response system.¶

Step 1: The impulse response h(τ)h(\tau) is flipped or time-reversed (that is, reflected about the origin) to obtain h(−τ)h(-\tau) and then shifted by to form h(t−τ)=h[−(τ−t)]h(t - \tau) = h[-(\tau - t)], which is a function of τ\tau with parameter tt.¶

Step 2-3 The signal x(τ)x(\tau) and h(t−τ)h(t - \tau) are multiplied together for all values of τ\tau with tt fixed at some value. Then, the product x(τ)h(t−τ)x(\tau)h(t - \tau) is integrated over all τ\tau to produce a single output value y(t)y(t).¶

Drill Problem 1: Compute the convolution, (x∗h)(t)(x \ast h)(t), of the signals represented in the figure above:¶

Step 1: The impulse response h(τ)h(\tau) is flipped or time-reversed (that is, reflected about the origin) to obtain h(−τ)h(-\tau) and then shifted by to form h(t−τ)=h[−(τ−t)]h(t - \tau) = h[-(\tau - t)], which is a function of τ\tau with parameter tt.¶

Step 2-3 The signal x(τ)x(\tau) and h(t−τ)h(t - \tau) are multiplied together for all values of τ\tau with tt fixed at some value. Then, the product x(τ)h(t−τ)x(\tau)h(t - \tau) is integrated over all τ\tau to produce a single output value y(t)y(t).¶

Case I: NO Overlap, y(t)=0y(t) = 0 at −∞→t1-\infty \to t_1, where t1=−1t_1 = -1¶

Case II: Leading-edge overlap, at t1→t2t_1 \to t_2 or at −1<t<0 -1 < t < 0 ¶

Case III: Full overlap, at t=0t = 0 to t=1t = 1¶

Case IV: Trailing edge overlap, at 1<t<21 < t < 2¶

Case V: No overlap, at t>2 t > 2¶

1. Final Answer¶

Drill Problem 2: Compute the convolution, (h∗x)(t)(h \ast x)(t), of the signals represented in the figure above:¶

Step 1: The impulse response h(τ)h(\tau) is flipped or time-reversed (that is, reflected about the origin) to obtain h(−τ)h(-\tau) and then shifted by to form h(t−τ)=h[−(τ−t)]h(t - \tau) = h[-(\tau - t)], which is a function of τ\tau with parameter tt.¶

Step 2-3 The signal x(τ)x(\tau) and h(t−τ)h(t - \tau) are multiplied together for all values of τ\tau with tt fixed at some value. Then, the product x(τ)h(t−τ)x(\tau)h(t - \tau) is integrated over all τ\tau to produce a single output value y(t)y(t).¶

Case I: NO Overlap, y(t)=0y(t) = 0 at −∞→t1-\infty \to t_1, where t1=0t_1 = 0¶

Case II: Leading-edge overlap, at t1→t2t_1 \to t_2 or at 0<t<1 0 < t < 1 ¶

Case III: Full overlap, at t=1t = 1¶

Case IV: Trailing edge overlap, at 1<t<21 < t < 2¶

Case V: No overlap, at t>2 t > 2¶

2. Final Answer¶

A convolution, $(h*x)(t)$ , is a mathematical operator which manipulates two functions $x$ and $h$ and produces an output that represents the amount of overlap between $x$ and a reversed and translated version of $h$ .

In signal processing, one of the functions ( $h$ ) is taken to be a fixed filter impulse response, and is known as a kernel.¶

1. The impulse response $h(\tau)$ is flipped or time-reversed (that is, reflected about the origin) to obtain $h(-\tau)$ and then shifted by to form $h(t - \tau) = h[-(\tau - t)]$ , which is a function of $\tau$ with parameter $t$ .¶

2. The signal $x(\tau)$ and $h(t - \tau)$ are multiplied together for all values of $\tau$ with $t$ fixed at some value.¶

3. The product $x(\tau)h(t - \tau)$ is integrated over all $\tau$ to produce a single output value $y(t)$ .¶

Step 1: The impulse response $h(\tau)$ is flipped or time-reversed (that is, reflected about the origin) to obtain $h(-\tau)$ and then shifted by to form $h(t - \tau) = h[-(\tau - t)]$ , which is a function of $\tau$ with parameter $t$ .¶

Step 2-3 The signal $x(\tau)$ and $h(t - \tau)$ are multiplied together for all values of $\tau$ with $t$ fixed at some value. Then, the product $x(\tau)h(t - \tau)$ is integrated over all $\tau$ to produce a single output value $y(t)$ .¶

Drill Problem 1: Compute the convolution, $(x \ast h)(t)$ , of the signals represented in the figure above:¶

Step 1: The impulse response $h(\tau)$ is flipped or time-reversed (that is, reflected about the origin) to obtain $h(-\tau)$ and then shifted by to form $h(t - \tau) = h[-(\tau - t)]$ , which is a function of $\tau$ with parameter $t$ .¶

Step 2-3 The signal $x(\tau)$ and $h(t - \tau)$ are multiplied together for all values of $\tau$ with $t$ fixed at some value. Then, the product $x(\tau)h(t - \tau)$ is integrated over all $\tau$ to produce a single output value $y(t)$ .¶

Case I: NO Overlap, $y(t) = 0$ at $-\infty \to t_1$ , where $t_1 = -1$ ¶

Case II: Leading-edge overlap, at $t_1 \to t_2$ or at $-1 < t < 0$ ¶

Case III: Full overlap, at $t = 0$ to $t = 1$ ¶

Case IV: Trailing edge overlap, at $1 < t < 2$ ¶

Case V: No overlap, at $t > 2$ ¶

Drill Problem 2: Compute the convolution, $(h \ast x)(t)$ , of the signals represented in the figure above:¶

Step 1: The impulse response $h(\tau)$ is flipped or time-reversed (that is, reflected about the origin) to obtain $h(-\tau)$ and then shifted by to form $h(t - \tau) = h[-(\tau - t)]$ , which is a function of $\tau$ with parameter $t$ .¶

Step 2-3 The signal $x(\tau)$ and $h(t - \tau)$ are multiplied together for all values of $\tau$ with $t$ fixed at some value. Then, the product $x(\tau)h(t - \tau)$ is integrated over all $\tau$ to produce a single output value $y(t)$ .¶

Case I: NO Overlap, $y(t) = 0$ at $-\infty \to t_1$ , where $t_1 = 0$ ¶

Case II: Leading-edge overlap, at $t_1 \to t_2$ or at $0 < t < 1$ ¶

Case III: Full overlap, at $t = 1$ ¶

Case IV: Trailing edge overlap, at $1 < t < 2$ ¶

Case V: No overlap, at $t > 2$ ¶