1. The Laplace transform of f(t)= cos(1.5t)e−0.5tu0(t) is given by
s+0.5
F(s)=
(s+0.5)2+2.25
Solve for the time response of a spring satisfying
x¨+2.25x = f(t)
with x(0)= 0 and ˙x = 0 for t going from zero to 50 seconds. Use system.impulse to do the computation
2. Solve the above problem with a much smaller decay:
f(t)= cos(1.5t)e−0.05tu0(t)
3. Consider the problem to be an LTI system. f(t) is the input, and x(t) is the output. Obtain the system transfer function X(s)/F(s). Now use signal.lsim to simulate the problem. In a for loop, vary the frequency of the cosine in f(t) from 1.4 to 1.6 in steps of 0.05 keeping the exponent as exp(−0.05t) and plot the resulting responses. Explain what is happening.
4. Solve for a coupled spring problem:
x¨+(x−y) = 0 y¨+2(y−x) = 0
where the initial condition is x(0)= 1, ˙x(0)= y(0)= y˙(0)= 0. Substitute for y from the first equation into the second and get a fourth order equation. Solve for its time evolution, and from it obtain x(t) and y(t) for 0 ≤ t ≤ 20.
5. Obtain the magnitude and phase response of the Steady State Transfer function of the following two-port network.
6. Consider the problem in Q5. suppose the input signal vi(t) is given by
vi
Obtain the output voltage v0(t) by defining the transfer function as a system and obtaining the output using signal.lsim.
How do you explain the output signal for 0 < t < 30µs? Can you explain the long term response on the msec timescale?
Note: You need to capture the fast variation, which means your time step should be smaller than 10−6. Yet you need to see the slow time, which means you must simulate till about 10 msec. Use an appropriate time vector.
Note: Since the network is resistive, the current at t = 0− will have decayed to zero, which gives you your initial condition.
