Consider the following transfer functions:

1. \(\frac{1}{{j\omega + 1}}\)

2. \(\frac{1}{{{{\left( {j\omega + 1} \right)}^2}}}\) 

3. \(\frac{1}{{\left( {j\omega + 1} \right)\left( {j\omega + 2} \right)}}\) 

The transfer functions which have a linear phase are:

Concept:

Linear phase

FIR filters can satisfy a linear phase response if the impulse response is symmetric or anti-symmetric.

• For symmetric impulse response phase will be θ(ω) = - α ω
• For anti-symmetric impulse response will be θ(ω) = - α × ω + β

Calculation:

Given system 1 is

\(G\left( s \right) = \frac{1}{{j\omega + 1}}\)

The phase of this system is

\(\angle G\left( s \right) = 0^\circ - {\tan ^{ - 1}}\omega \)

θ₁(ω) = - α ω

So this is the linear phase.

Given system 2 is

\(G\left( s \right) = \frac{1}{{{{\left( {j\omega + 1} \right)}^2}}}\) 

The phase of this system is

\(\angle G\left( s \right) = 0^\circ - 2{\tan ^{ - 1}}\omega \)

θ₂(ω) = - α ω

So this is the linear phase.

Given system 3 is

\(G\left( s \right) = \frac{1}{{\left( {j\omega + 1} \right)\left( {j\omega + 2} \right)}}\) 

The phase of this system is

\(\angle G\left( s \right) = 0^\circ - {\tan ^{ - 1}}\omega - {\tan ^{ - 1}}\frac{\omega }{2}\) 

\(\angle G\left( s \right) = - {\tan ^{ - 1}}\left( {\frac{{\omega + \frac{\omega }{2}}}{{1 - \frac{{\omega \times \omega }}{2}}}} \right)\) 

θ₃(ω) = - α ω

So this is the linear phase.

All are linear phase systems