A system has transfer function H(s) = (s + 2) / ((s + 1)(s + 3)). Which values of s are the poles of this system?
As = −2 only
Bs = −1 and s = −3
Cs = 1 and s = 3
Ds = −2 and s = 0
Poles are the roots of the denominator polynomial. Setting (s+1)(s+3) = 0 gives s = −1 and s = −3. The zero (root of the numerator s+2 = 0) is at s = −2. A common confusion is swapping poles and zeros; remember: denominator → poles, numerator → zeros.
Question 2 True / False
A system whose transfer function has a pole at s = 2 + 3j (positive real part) will produce a bounded, stable output for any bounded input.
TTrue
FFalse
Answer: False
Poles with positive real parts correspond to exponentially growing components e^(σt) with σ > 0. The output grows without bound for any non-zero initial condition or input, making the system unstable. BIBO (bounded-input bounded-output) stability requires ALL poles to lie in the open left half of the s-plane, i.e., Re(s) < 0 for every pole.
Question 3 Short Answer
Why is the transfer function H(s) = Y(s)/X(s) useful for analyzing LTI systems, rather than working directly with convolution in the time domain?
Think about your answer, then reveal below.
Model answer: Convolution in the time domain becomes multiplication in the s-domain. H(s) encodes the system's complete input-output behavior as a single rational function, so the output for any input X(s) is just Y(s) = H(s)·X(s). This algebraic multiplication is far simpler than computing a convolution integral, and the pole-zero structure directly reveals stability, natural frequencies, and frequency response without solving differential equations.
The convolution theorem for the Laplace transform is the mathematical reason: if y(t) = h(t)*x(t) then Y(s) = H(s)X(s). The transfer function is essentially a compressed description of the system's impulse response in a form that makes composition of systems (cascades, feedback loops) trivial to analyze.