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Three blocks having transfer functions \(G_1(s) = \frac{1}{s+2}\), \(G_2(s) = \frac{1}{s+5}\) and \(G_3(s) = \frac{s+1}{s+3}\) are cascaded. The equivalent transfer function is:

This question was previously asked in
UGC NET 2023 Electronic Science Question Paper (13-Dec-2023) (Shift 1)
The correct answer is

\(\frac{s+1}{(s+2)(s+3)(s+5)}\)

Rule for cascaded blocks. When blocks are connected in series (cascade) and there is no loading between them — the standard assumption in block-diagram algebra — the output of one is the input of the next, so the transfer functions simply multiply:

\(G(s)=G_1(s)\,G_2(s)\,G_3(s)\)

Substitute the three blocks.

\(G(s)=\dfrac{1}{s+2}\times\dfrac{1}{s+5}\times\dfrac{s+1}{s+3}\)

Multiply numerators and denominators.

\(G(s)=\dfrac{s+1}{(s+2)(s+3)(s+5)}\)

Reading the result. The overall system has poles at s = −2, −3, −5 (all in the left half plane, so it is stable) and a zero at s = −1. Cascading therefore collects the poles and zeros of the individual blocks — a useful check: the number of poles of the cascade equals the total number of poles of the parts.

Why the other options are wrong.

The negative signs in options 2 and 3 could only come from a negative feedback loop or an inverting summing junction; a plain cascade introduces no sign change.

The numerator \(s^{3}+10s^{2}+30s+31\) in options 3 and 4 is what you get by adding the three transfer functions over a common denominator — that is the rule for blocks connected in parallel (all fed by the same input, outputs summed), not in cascade. Recognising the difference between "series → multiply" and "parallel → add" is the whole point of the question.

Practical caveat. The multiplication rule assumes each stage has zero output impedance and infinite input impedance. Two RC sections wired directly together do not multiply, because the second stage loads the first; that is exactly why buffer amplifiers are inserted between filter stages.

Hence, the equivalent transfer function is \(\dfrac{s+1}{(s+2)(s+3)(s+5)}\), i.e. option 1.

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