Preliminary¶

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import numpy as np
import matplotlib.pyplot as plt
%pylab inline --no-import-all

Populating the interactive namespace from numpy and matplotlib

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def plotSignal(t,x,ptitle='Signal'): 
    fig = plt.figure(figsize=(10,6))    
    ax1 = fig.add_axes([0, 0, 1, 0.8])   # [left, bottom, width, height]
    ax1.set_xlabel('t')
    ax1.set_ylabel('x')
    ax1.set_title(ptitle)
    ax1.fill_between(t,0,x,color='green', alpha=0.4) 
    ax1.plot(t, x, color='green', lw=6);

1. Express the voltage waveform $v(t)$ shown below, as a sum of unit step functions for the time interval $0 < t < 7$ s.¶

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

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t = np.linspace(0, 7, 1000, endpoint=True)

# Time breakpoints
t0 = 0
t1 = 2
t2 = 3
t3 = 5
t4 = 7

# Step functions
u0 = step(t, t0)
u1 = step(t, t1)
u2 = step(t, t2)
u3 = step(t, t3)
u4 = step(t, t4)

# Pulse Signals
pu1 = u0 - u1
pu2 = u1 - u2
pu3 = u2 - u3
pu4 = u3 - u4

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# First Pulse
fig = plt.figure(figsize=(16,2))
plt.subplot(1,4,1)
plt.plot(t,pu1,'g-')
plt.fill_between(t,0,pu1,color='green', alpha=0.4) 
plt.xlabel('t ($s$)')
plt.ylabel('v ($t$)') 
plt.title('$u_0 - u_1$')

# Second Pulse
plt.subplot(1,4,2)
plt.plot(t,pu2,'g-')
plt.fill_between(t,0,pu2,color='green', alpha=0.4) 
plt.xlabel('t ($s$)')
plt.ylabel('v ($t$)');
plt.title('$u_1 - u_2$')

# Third Pulse
plt.subplot(1,4,3)
plt.plot(t,pu3,'g-')
plt.fill_between(t,0,pu3,color='green', alpha=0.4) 
plt.xlabel('t ($s$)')
plt.ylabel('v ($t$)');
plt.title('$u_2 - u_3$')

# Fourth Pulse
plt.subplot(1,4,4)
plt.plot(t,pu4,'g-')
plt.fill_between(t,0,pu4,color='green', alpha=0.4) 
plt.xlabel('t ($s$)')
plt.ylabel('v ($t$)');
plt.title('$u_3 - u_4$');

Notice the unit step signals are acting as a 'switch'¶

Solution:¶

Signal at time, $t = 0$ to $t = 2$ :¶

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s1 = 20*np.exp(-2*t)*pu1

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plotSignal(t,s1)

Signal at time, $t = 2$ to $t = 3$ :¶

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s2 = (10*t - 30) * pu2

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plotSignal(t,s2)

Signal at time, $t = 3$ to $t = 5$ :¶

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s3 = (-10*t + 50) * pu3

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plotSignal(t,s3)

Signal at time, $t = 5$ to $t = 7$ :¶

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s4 = (10*t - 70) * pu4

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plotSignal(t, s4)

Total Signal¶

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s_t = s1 + s2 + s3 + s4

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plotSignal(t, s_t)

2. Simplify the ff. expressions:¶

$\text{a.} \: \int_{-5}^1 \, {7e^{t}}^2 \, \cos(t) \delta (t - 2) dt = 0 \qquad \text{: Sifting Property}$

Note: the limit from -5 to 1 doesn't include $t_0 = 2$

$\text{b.} \: \int_{- \infty}^{t-1} \delta(\tau + 3) e^{\tau} d\tau \qquad \text{: Sifting Property}$

$= e^{-3} \int_{- \infty}^{t-1} \delta(\tau + 3) d\tau$

$= e^{-3} u(\tau + 3)\Bigg|_{- \infty}^{t-1}$

$= e^{-3} u(t - 1 + 3) - e^{-3} u(- \infty)$

$\int_{- \infty}^{t-1} \delta(\tau + 3) e^{\tau} d\tau = e^{-3} u(t + 2)$

3. Determine whether the ff. signal is odd/even/neither: ____________¶

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def triangular(t,tau):
    x = np.zeros(len(t))
    for k,tk in enumerate(t):
        if np.abs(tk) <= tau/1.:
            x[k] = 1 - np.abs(tk)/tau
        else:
            x[k] = 0
    return x

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t = np.linspace(-3, 3, 1000, endpoint=True)

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tri_1 = triangular(t-1,1)

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plotSignal(t,tri_1)

$x(t) \ne x(-t) \qquad \text{(Not Even)}$

$x(t) \ne -x(-t) \qquad \text{ (Not Odd)}$

4. Sketch and label the ff:¶

$x(t) = \sqrt{2} \left( 1 + j \right) e^{j\pi/4} e^{\left( -1 + j 2\pi \right)t}$

a.) Re{x(t)}
b.) Im{x(t)}

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t = np.linspace(0, 2*np.pi, 1000, endpoint=True)
x_t = np.sqrt(2) * (1+1j) * np.exp((np.pi/4)*1j) * np.exp((-1 + 2*np.pi*1j)*t)

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x_real = np.real(x_t)

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x_imag = np.imag(x_t)

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plotSignal(t,x_real)

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plotSignal(t,x_imag)

5. Let $x(t)$ and $y(t)$ be periodic signals with fundamental periods $T_1$ and $T_2$ , respectively. Under what conditions is the sum $x(t) + y(t)$ periodic, and what is the fundamental period of this signal if it is periodic?¶

The sum $x(t) + y(t)$ is periodic if $x(t)$ and $y(t)$ have a common period. That is, $nT_1 = mT_2$ .

6. Determine whether the ff. signals are periodic:¶

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t = np.linspace(0, 12*np.pi, 10000, endpoint=True)
x1 = np.exp(np.sin(t))
plotSignal(t,x1)

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t = np.linspace(0, 12*np.pi, 10000, endpoint=True)
x2 = np.exp(t*np.sin(t))
plotSignal(t,x2)

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t = np.linspace(0, 12*np.pi, 10000, endpoint=True)
x3 = np.sin(6*np.pi*t/7) + 2*np.cos(3*t/5)
plotSignal(t,x3)

Aperiodic Signal¶

7. What type of signals can be represented as sums of sinusoids:
Energy or Power? ___________¶

Power Signal

8. Sketch and label each of the following signals:¶

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t = np.linspace(-2, 3, 1000, endpoint=True)

# Time breakpoints
t1 = -1
t2 = 0
t3 = 1
t4 = 2

# Step functions
u1 = step(t,-1)
u2 = step(t,0)
u3 = step(t,1)
u4 = step(t,2)

# Pulse Signals
pu1 = u1 - u2
pu2 = u2 - u3
pu3 = u3 - u4

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# First Pulse
fig = plt.figure(figsize=(16,2))
plt.subplot(1,3,1)
plt.plot(t,pu1,'g-')
plt.fill_between(t,0,pu1,color='green', alpha=0.4) 
plt.xlabel('t ($s$)')
plt.ylabel('v ($t$)') 
plt.title('$u_1 - u_2$')

# Second Pulse
plt.subplot(1,3,2)
plt.plot(t,pu2,'g-')
plt.fill_between(t,0,pu2,color='green', alpha=0.4) 
plt.xlabel('t ($s$)')
plt.ylabel('v ($t$)');
plt.title('$u_2 - u_3$')

# Third Pulse
plt.subplot(1,3,3)
plt.plot(t,pu3,'g-')
plt.fill_between(t,0,pu3,color='green', alpha=0.4) 
plt.xlabel('t ($s$)')
plt.ylabel('v ($t$)');
plt.title('$u_3 - u_4$');

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s = (t+1)*pu1 + pu2 + 2*pu3
plotSignal(t, s)

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xa = s * step(1-t,0)
plotSignal(t,xa)

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xb = s * (step(t,0) - step(t-1,0))
plotSignal(t,xb)

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