GATE 2009 Signals and Systems Questions and Solutions

An LTI system with impulse response \(h(t)\) produces output \(y(t)\) for input \(x(t)\). When the input \(x(t-\tau)\) is applied to a system with impulse response \(h(t-\tau)\), the output is:

1. \(y(t)\)
2. \(y(2(t-\tau))\)
3. \(y(t-\tau)\)
4. \(y(t-2\tau)\)

Solution

Originally \(Y(s)=H(s)X(s)\). Shifting input and impulse response each by \(\tau\) multiplies their transforms by \(e^{-s\tau}\):

Taking the inverse transform, \(y'(t) = y(t-2\tau)\).

A cascade of three LTI systems is causal and unstable. From this, we conclude that:

1. each system in the cascade is individually causal and unstable
2. at least one system is unstable and at least one system is causal
3. at least one system is causal and all systems are unstable
4. the majority are unstable and the majority are causal

Solution

For a cascade, \(h(t)=h_1(t)*h_2(t)*h_3(t)\). If the overall system is causal, the earliest nonzero time \(t_1+t_2+t_3\ge 0\), which requires at least one subsystem to be causal. If the cascade is unstable, at least one subsystem must be unstable (a cascade of stable systems is stable). The other options over-constrain the result.

The Fourier series coefficients of a periodic signal \(x(t)=\sum_{k} a_k e^{j2\pi k t/T}\) are \(a_{-2}=2-j,\ a_{-1}=0.5+j0.2,\ a_0=j2,\ a_1=0.5-j0.2,\ a_2=2+j\), and \(a_k=0\) for \(|k|>2\). Which is true?

1. \(x(t)\) has finite energy because only finitely many coefficients are non-zero
2. \(x(t)\) has zero average value because it is periodic
3. The imaginary part of \(x(t)\) is constant
4. The real part of \(x(t)\) is even

Solution

Group conjugate-index pairs. For \(k=\pm2\): \((2-j)e^{-j2\omega_0 t}+(2+j)e^{j2\omega_0 t} = 4\cos 2\omega_0 t - 2\sin 2\omega_0 t\). For \(k=\pm1\): \((0.5+j0.2)e^{-j\omega_0 t}+(0.5-j0.2)e^{j\omega_0 t} = \cos\omega_0 t + 0.4\sin\omega_0 t\). Adding \(a_0 = j2\):

The imaginary part is the constant \(2\). (The real part is neither even nor odd, so D is false.)

The Z-transform of \(x[n]\) is \(X(z)=4z^{-3}+3z^{-1}+2-6z^{2}+2z^{3}\). It is applied to a system with \(H(z)=3z^{-1}-2\) giving output \(y[n]\). Which is true?

1. \(y[n]\) is non-causal with finite support
2. \(y[n]\) is causal with infinite support
3. \(y[n]=0\) for \(|n|>3\)
4. \(\operatorname{Re}[Y(z)]_{z=e^{j\theta}} = -\operatorname{Re}[Y(z)]_{z=e^{-j\theta}}\)

Solution

The output transform is the product

Expanding gives terms ranging from \(z^{-4}\) up to \(z^{3}\):

The positive powers of \(z\) mean \(y[n]\ne 0\) for some \(n<0\) (non-causal), while only finitely many terms appear (finite support).