Question

A transfer function is given by \( H(s) = \frac{2(s+2)(s+4)}{(s+3)(s+5)(s+7)} \). Then the transfer function becomes zero when

• A. \( s=0 \)
• B. \( s=-2 \)
• C. \( s=-3 \)
• D. \( s=-5 \)

Hint:

Zeros of \(H(s)\) are the values of \(s\) for which \(N(s)=0\) (numerator is zero).
Poles of \(H(s)\) are the values of \(s\) for which \(D(s)=0\) (denominator is zero).

Answer 1

The Correct Option is B

A transfer function \(H(s)\) becomes zero when its numerator is zero (and denominator is non-zero). 

The numerator of \( H(s) = \frac{2(s+2)(s+4)}{(s+3)(s+5)(s+7)} \) is \(N(s) = 2(s+2)(s+4)\).

Set \(N(s) = 0 \Rightarrow 2(s+2)(s+4) = 0\).

This gives \(s+2=0\) or \(s+4=0\).

So, \(s=-2\) or \(s=-4\). These are the zeros of \(H(s)\).

At \(s=-2\), denominator \(= (-2+3)(-2+5)(-2+7) = (1)(3)(5) = 15 \neq 0\).

So \(H(-2) = 0/15 = 0\).

At \(s=-3\) or \(s=-5\), the denominator is zero, so these are poles.

At \(s=0\), \(H(0) = \frac{2(2)(4)}{3 \cdot 5 \cdot 7} = \frac{16}{105} \neq 0\).

Thus, the transfer function becomes zero at \(s=-2\).

\[ \boxed{s=-2} \]