The transfer function \(H\left( s \right) = \frac{1}{{\left( {s +

SOLUTION

Analysis:

The given transfer function is:

\(H\left( s \right) = \frac{1}{{\left( {s + 1} \right)}}\)

Replacing s with jω, we get:

\(H\left( jω \right) = \frac{1}{{\left( { jω + 1} \right)}}\)

At ω = 0 (Low frequency)

\(H\left( 0 \right) = \frac{1}{{\left( { j0 + 1} \right)}}=1\)

We observe that the transfer function is unity for low frequencies, i.e. the input signal at low frequencies is easily passed to the output without much attenuation.

At ω → ∞ (High frequencies)

\(H\left(j\omega \right)_{\omega \rightarrow \infty} = \frac{1}{{\left( { j\omega + 1} \right)}}|_{\omega \rightarrow \infty }=0\)

We observe that the circuit is not allowing the high frequencies to pass, i.e. the high frequencies signals are attenuated.

∴ The given circuit is a low pass filter.

The circuit diagram of a passive low pass filter is as shown:

The transfer function is the ratio of the output voltage to the input voltage, i.e.

 \(\frac{{{V_0}}}{{{V_1}}} = \frac{{\frac{1}{{j\omega C}}}}{{R + \frac{1}{{j\omega C}}}} \)

\(H(j\omega)= \frac{1}{{1 + j\omega CR}}\)