Question

TASKS Consider the LTI systems defined by following impulse responses

S₁​:h(t)=u(t)−u(t−5);
S₂​:h(t)=(−t+2)[u(t)−u(t−4)];
S₃​:h(t)=e⁻/⁶u(t);
S₄​:h(t)=et/2u(t);
S₅​:h[n]=u[n+1]−u[n−3];​
For these systems, "convolution integral" or "convolution sum" will be implemented. Do not use the "conv" command of MATLAB. 1) For the systems S1​ and S2​, assuming x(t)=5[u(t)−u(t−2)] is the input for both systems, calculate the outputs y(t). Verify your results by implementing convolution using MATLAB. Comment on the causality and stability of the systems.

Answer 1

The student is asked to calculate the outputs of two continuous-time LTI systems by performing the convolution integral and to verify the results using MATLAB. A proper assessment of causality and stability for each system is also required.

Step-by-step explanation:

The question pertains to calculating the outputs of two Linear Time-Invariant (LTI) systems S1 and S2 given their impulse responses and a specific input signal. To find the system outputs y(t), we implement the convolution integral for continuous-time signals since the continuous input x(t) = 5[u(t) - u(t-2)] is specified.

Calculation Steps for S1 and S2:

For system S1, with h(t) = u(t) - u(t-5), we perform the convolution of h(t) and x(t) manually by integrating piecewise over the relevant intervals.

For system S2, with h(t) = (-t + 2)[u(t) - u(t - 4)], we again perform the convolution of h(t) and x(t), considering the intervals affected by the unit step functions.

After manually calculating the convolutions, we then verify results using MATLAB (excluding the use of the 'conv' command). We also analyze the causality and stability of systems S1 and S2 based on their impulse responses.

Comment on Causality and Stability:

Causality can be discussed by looking at the impulse response and checking if the system's output depends only on past and present inputs.

Stability can be assessed by checking if the impulse response is absolutely integrable, i.e., the integral of the absolute value of h(t) over all time is finite.