Chapter 5_Signal Waveforms

# Chapter 5_Signal Waveforms - The Analysis and Design of...

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The Analysis and Design of Linear Circuits Seventh Edition 5 Signal Waveforms 5.1 Exercise Solutions Exercise 5–1. Write an expression using unit step functions for the waveform in Figure 5–6. The signal has a positive transition with a magnitude of 10 at t = - 2 s, a negative transition with a magnitude of 15 at t = 2 s, and a positive transition with a magnitude of 5 at t = 4 s. The following expression describes the waveform. v ( t ) = 10 u ( t + 2) - 15 u ( t - 2) + 5 u ( t - 4) V Exercise 5–2. Figure 5–9 purports to be an alternative description of an impulse function as ε 0. Prove or disprove the claim. From the figure, the area of the triangle is 1 2 (2 ε )(1 ) = 1. As ε 0, the base of the triangle shrinks to zero but the amplitude grows to infinity. The area always remains at one. Hence, this is equivalent to the definition of an impulse and proves the claim. Exercise 5–3. Write an expression using ramp functions to describe the waveform shown in Figure 5–12. The waveform has a ramp function that starts at t = - 1 s and has a slope of 2 / (1 - ( - 1)) = +1. The waveform then changes from a slope of +1 to - 1 at t = 1 s. Finally, at t = 3 s, the slope changes from - 1 to zero. The following expression describes the waveform. v ( t ) = r ( t + 1) - 2 r ( t - 1) + r ( t - 3) V Exercise 5–4. Express the following signals in terms of singularity functions: (a). v 1 ( t ) = 0 t < 2 4 2 < t < 4 - 4 4 < t The signal transitions from 0 to 4 at t = 2 and then transitions from 4 to - 4 at t = 4. We can create this signal with two step functions. v 1 ( t ) = 4 u ( t - 2) - 8 u ( t - 4) (b). v 2 ( t ) = 0 t < 2 4 2 < t < 4 - 2 t + 12 4 < t The signal transitions from 0 to 4 at t = 2 and then transitions to have a slope of - 2 at t = 4. We can create this signal with a step function and a ramp function. v 2 ( t ) = 4 u ( t - 2) - 2 r ( t - 4) (c). v 3 ( t ) = integraldisplay t −∞ v 1 ( x ) dx The integral of a step function is a ramp function. We can write v 3 ( t ) directly by inspection. v 3 ( t ) = 4 r ( t - 2) - 8 r ( t - 4) Solution Manual Chapter 5 Page 5-1

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The Analysis and Design of Linear Circuits Seventh Edition (d). v 4 ( t ) = dv 2 ( t ) dt The derivative of a step function is an impulse function and the derivative of a ramp function is a step function. We can write v 4 ( t ) directly by inspection. v 4 ( t ) = 4 δ ( t - 2) - 2 u ( t - 4) Exercise 5–5. Sketch the waveform described by v ( t ) = 20 e 10000 t u ( t ) V The waveform starts at t = 0 with a magnitude of 20 V. The waveform then exponentially decays with a time constant of T C = 1 / 10000 = 100 μ s. The following MATLAB code plots the waveform. % Create a time vector t = -100e-6:100e-9:500e-6; % Create the signal vt = 20 * exp(-10000 * t). * heaviside(t); % Plot the results figure plot(t,vt, 'b' , 'LineWidth' ,3) grid on axis([-100e-6,500e-6,0,20]); xlabel( 'Time, (s)' ) ylabel( 'v(t), (V)' ) The corresponding MATLAB output is shown below. -1 0 1 2 3 4 5 x 10 -4 0 2 4 6 8 10 12 14 16 18 20 Time, (s) v(t), (V) Exercise 5–6. Figure 5–20 contains three exponential waveforms. Match each curve with the appropriate expression.
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• Fall '15
• Derivative, Exponential decay, Waveform, Solution Manual Chapter

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