Q.43 The output y(t) of a first-order process is governed by the following differential equation:
τ_pdy/dt + y = K_p f(t)

where τ_p is a non-zero time constant, K_p is the gain and
f(t) is the input with f(0) = 0.

Assume y(0) = 0. The transfer function for this process
(consider s as the independent variable in the Laplace domain) is:

(A) K_p / (τ_ps + 1)

(B) τ_p / (K_ps + 1)

(D) K_p / (τ_p(s + 1))

First-Order Process Transfer Function

The correct transfer function for the given first-order process
is option (A):

G(s) =
K_p /
(τ_ps + 1)

Derivation Process

Start with the first-order differential equation:

τ_p dy/dt + y = K_p f(t)

Assume zero initial conditions:

Applying the Laplace transform:

τ_p sY(s) + Y(s) = K_p F(s)

Factorizing:

Y(s)(τ_ps + 1) = K_p F(s)

Hence, the transfer function becomes:

G(s) = Y(s)/F(s) = K_p / (τ_ps + 1)

(A) K_p / (τ_ps + 1):
Correct. Matches the standard first-order form where
steady-state gain is K_p and
time constant is τ_p.
(B) τ_p / (K_ps + 1):
Incorrect. Gain becomes τ_p, not K_p, and
parameters are mismatched.
(C) τ_p / (K_p(s + 1)):
Incorrect structure. Wrong numerator–denominator relationship and
improper handling of τ_p.
(D) K_p / (τ_p(s + 1)):
Close but incorrect. Effectively changes the time constant to 1,
not τ_p.

First-order transfer functions commonly model systems such as
tanks, heaters, and thermal processes in control engineering.

The pole at:

s = −1 / τ_p

determines the exponential decay rate of the system response.

The corresponding step response is:

y(t) = K_p(1 − e^−t/τ_p)u(t)