Question

If the number of zeros is less than the number of poles, i.e. \( Z<P \), then the value of the transfer function becomes zero for \( s \to \infty \). Hence, some zeros are located at infinity and the order of such zeros will be:

• A. Z
• B. P + Z
• C. Z - P
• D. P - Z

Hint:

The number of zeros at infinity for a rational transfer function is simply the difference between the degree of the denominator (number of poles, P) and the degree of the numerator (number of zeros, Z). This only applies when \(P>Z\).

Answer 1

The Correct Option is D

Step 1: Define a general transfer function. A rational transfer function \(H(s)\) can be written as the ratio of two polynomials: \[ H(s) = K \frac{N(s)}{D(s)} = K \frac{s^Z + b_{Z-1}s^{Z-1} + \dots + b_0}{s^P + a_{P-1}s^{P-1} + \dots + a_0} \] Here, Z is the number of finite zeros (the degree of the numerator \(N(s)\)) and P is the number of finite poles (the degree of the denominator \(D(s)\)).
Step 2: Analyze the behavior as \( s \to \infty \). To find the limit of \(H(s)\) as \( s \to \infty \), we look at the terms with the highest power of \(s\) in the numerator and denominator: \[ \lim_{s \to \infty} H(s) = \lim_{s \to \infty} K \frac{s^Z}{s^P} = \lim_{s \to \infty} K s^{Z-P} \]
Step 3: Apply the given condition \( Z<P \). The condition \(Z<P\) means that \(Z - P\) is a negative number. Let \(Z - P = -N\), where \(N = P - Z\) is a positive integer. The limit becomes: \[ \lim_{s \to \infty} H(s) = \lim_{s \to \infty} K s^{-N} = \lim_{s \to \infty} \frac{K}{s^N} \] As \(s\) approaches infinity, \(s^N\) also approaches infinity (since N>0), and therefore \(K/s^N\) approaches 0.
Step 4: Define zeros at infinity. When the transfer function approaches zero as \(s \to \infty\), it is said to have zeros at infinity. The order (or number) of these zeros is the value N such that \(H(s)\) behaves like \(1/s^N\). From our analysis, \(N = P - Z\). Therefore, the order of the zeros at infinity is \(P - Z\).