EE 2. Signals and Systems Solutions of homework 2 Spring 2 Exercise Due Date Week of 22 nd Feb. Problems Q Compute and sketch the output y[n] of each discrete-time LTI system below with impulse response h[n] and input signal x[n]. Use the graphical method to compute the discrete-time convolutions. (a) The impulse response h[n] and input signal x[n] are as depicted below. x[n]...... - 2 4 n h[n]... - 2 4... n The output of the system will be the sum of system response to signal h[], h[] and h[2]. The three signals are: h[]x[n] x[n], h[]x[n ] x[n ], h[2]x[n 2] x[n 2]. Thus, the output y[n] of the system will be y[n] [,,, 6, 7, 7, 4,,, ] ( indicates the n sample) (b) x[n] δ[n] + 2δ[n ] + δ[n 2], h[n] δ[n + ] δ[n] + δ[n ] LUMS School of Science & Engineering, Lahore, Pakistan.
y[n] [,,,,,,,, ] ( indicates the n sample) Q2 Compute the convolutions y[n] x[n] h[n] (a) x[n] α n u[n], h[n] β n u[n], α β. Sketch the output signal y[n] for the case α.8, β.9. y[n] x[k]h[n k] α k u[k]β n k u[n k] n ( ) k α β n, n β k ( ) n+ α β n β ( ) α u[n] β ( β n+ α n+ ) u[n], α β β α you can plug in the values of α and β to plot y[n] (b) x[n] δ[n] δ[n ], h[n] u[n] y[n] u[n] (c) x[n] u[n], h[n] u[n] x[k]h[n k] (δ[k]u[n k]) δ[k] u[n ] u[n] u[n ] δ[n] (δ[k] δ[k ])u[n k] (δ[k ]u[n k]) δ[k ] 2
y[n] x[k]h[n k] u[k]u[n k] (u[k]u[ (k n)]) { n k n +, n, n < (n + )u[n] (d) The input signal and impulse response depicted below. Sketch the output signal y[n] x[n] - 2 4 n h[n] -9-8 -7-6 -5-4 - -2-2 4 5 6 7 8 9 n y[n] [,,,,, 2,,,,, 2,,,, 2,,,,, 2,,, ] Q Compute and sketch the output y(t) of the continuous-time LTI system S with: Impulse response h(t) e t u(t ) Input signal x(t) u(t + ) u(t ), t < y(t) e t, t < 4 (e 4 )e t, t 4 Q4 Compute the response y(t) of a coninuous-time LTI system described by its impulse response h(t) e ( +j)t u(t) to the input signal x(t) u(t). Is the system BIBO stable? Is it causal?
The step response of the system is for t < and for t > is given by: y(t) Hence, y(t) h(t τ)x(τ)dτ t (e 2(t 4) e 2t ) (e 8 )e 2t j ( e ( +j)t )u(t) The system is BIBO stable since + h(t) + It is also causal since h(t), t < e ( +j)(t τ) dτ e t [ e t ] + j e ( +j)(t τ) t, t > < +. Q5 Consider the following first-order, causal LTI differential system S initially at rest: S : dy(t) + ay(t) dx(t) 2x(t), a > is real (a) Calculate the impulse response h (t) of the system S. Sketch it for a 2. (b) Is the system S BIBO stable? Justify your answer. Step : Set up the problem to calculate the impulse response of the left-hand side of the equation dh a (t) + ah a (t) δ(t) () Step 2: Find the initial condition of the corresponding homogeneous equation t + by integrating the above differential equation from t to t +. Note that the impulse will be in term dh a(t), so h a (t) will have a finite jump at most. Thus we have + dh a (τ) dτ dτ h a ( + ) hence h ( + ) is our initial condition for the homogeneous equation for t > dh a (t) + ah a (t) Step : The characteristic polynomial is p(s) s + a and it has one zero at s a, which means that the homogeneous response has the form h a (t) Ae at for t >. The initial condition allows us to determine the constant A: so that, h a ( + ) A, h a (t) e at u(t) 4
Now since LTI systems are commutative. This means that: dh a(t) h (t) 2h a (t) d ( e at u(t) ) 2e at u(t) (2 + a)e at u(t) + δ(t) The system is BIBO stable. The single zero of its characteristic polynomial is s a < Q6 Consider the following second-order, causal LTI differential system S 2 initially at rest: S 2 : d2 y(t) 2 + 5 dx(y) Calculate the impulse response h 2 (t) of the system S 2. + 6y(t) x(t) Step : Set up the problem to calculate the impulse response of the left-hand side of the equation d 2 h 2 (t) 2 + 5 dh 2(t) + 6h 2 (t) δ(t) (2) Step 2: Find the initial condition of the corresponding homogeneous equation t + by integrating the above differential equation from t to t +. Note that the impulse will be in term d2 h 2 (t) 2, so dh 2(t) will have a finite jump at most. Thus we have hence dh 2( + ) + d 2 h 2 (τ) dτ 2 dτ dh 2( + ) is one of our initial conditions for the homogeneous equation for t > d 2 h 2 (t) 2 + 5 dh 2(t) + 6h 2 (t) Since dh 2(t) has a finite jump at t to t +, the other initial condition is h 2 ( + ) ). Step : The characteristic polynomial is p(s) s 2 + 5s + 6 and it has zeros at s 2, s 2, which means that the homogeneous response has the form h a (t) Ae 2t +Be t for t >. The initial condition allows us to determine the constant A: so that, A, B and finally h 2 ( + ) A + B, dh 2 ( + ) 2A B h 2 (t) (e 2t e t )u(t) () 5
Q7 Consider the following second-order, causal difference LTI system S initially at rest: S : y[n] +.9y[n ] +.2y[n 2] x[n] (a) What is the characteristic polynomial of S? What are its zeros? p(z) z 2 +.9z +.2 (z +.4)(z +.5) The zeros are z.4, z 2.5. (b) Compute the impulse response of S for all n. The homogeneous response is given by y[n] A(.4) n + B(.5) n, n > The initial conditions for the homogeneous equation for n > are y[ ] and y[] δ[]. Now we can compute the coefficients A and B: y[ ] A(.4) + B(.5) 2.5A 2B y[] A + B Hence, and the impulse response is A 4, B 5 h [n] [ 4(.4) n + 5(.5) n ] u[n] 6