Question

For the system, \(y (t) = \int_{-∞}^{t+3} x(t) \,dt\), which of the following holds true?

(a) System is Linear, time-invariant and causal

(b) System is time-invariant and causal

(c) System is Linear and time-invariant

(d) System is Linear and stable

I had been asked this question by my school teacher while I was bunking the class.

Question is taken from Properties of Systems in chapter Signals and Systems Basics of Signals and Systems

Answer 1

Right answer is (c) System is Linear and time-invariant

The best I can explain: \(y_1 (t) = \int_{-∞}^{t+3} v(t) \,dt\)

And, \(y_2 (t) = \int_{-∞}^{t+3} kv(t) \,dt\)

= \(k \int_{-∞}^{t+3} x(t) \,dt\) = k y1 (t)

Now, x3 (t) = v (t) + w (t)

And, y3 (t) = \(\int_{-∞}^{t+3}\) [v(t) + w(t)]dt

= \(\int_{-∞}^{t+3}\)v(t)  dt + \(\int_{-∞}^{t+3}\)w(t)  dt

= y1 (t) + y2 (t)

Since, it is both homogeneous and additive, so linear system.

Again, y1 (t) = \(\int_{-∞}^{t+3}\) v(t)  dt

And, y2 (t) = \(\int_{-∞}^{t+3}\) v(t-t0)  dt

= y1 (t-t0)

∴ System is time invariant.

The response at any time t=t0, depends partially on the excitation at time t0 < t < (t0 + 3), which are in future, so non-causal system.

The response will increase without bound as time increases, so unstable system.