Frequency Response of the RC Circuit
Introduction
The frequency response of an RC circuit describes how the circuit’s impedance and phase change as the frequency of the applied alternating current (AC) signal varies. RC circuits, composed of a resistor (R) and a capacitor (C) in series, are foundational in analog electronics and signal processing. Their behavior changes significantly with frequency, making them essential in filters, signal conditioning, and frequency-sensitive applications.What Is Frequency Response?
Frequency response refers to a circuit’s behavior when an AC signal of varying frequency is applied. It shows how the magnitude and phase of the output or impedance change with frequency. In the case of a series RC circuit, the focus is often on how the total impedance and phase angle vary from low to high frequencies.How an RC Circuit Responds to Frequency
In a series RC circuit: $$ X_C = \frac{1}{2 \pi f C} $$ where:- (X_C) is capacitive reactance (Ω),
- (f) is the frequency (Hz),
- (C) is the capacitance (F).
Total Impedance of the RC Circuit
The total impedance ($Z_T$) of the series RC circuit combines the resistor and capacitor: $$ Z_T = R - jX_C $$ In polar form: $$ |Z_T| = \sqrt{R^2 + X_C^2} $$ $$ \theta = -\tan^{-1}\left(\frac{X_C}{R}\right) $$- At low frequencies, ($X_C$) is large, and the circuit behaves primarily like a capacitor — the impedance magnitude is high and the phase angle approaches ($-90^\circ$).
- At high frequencies, ($X_C$) becomes small and the resistor dominates — the impedance magnitude approaches (R) and the phase angle moves closer to ($0^\circ$).
Break (Corner) Frequency
A key frequency in the RC frequency response is the break frequency ($f_1$), where: $$ X_C = R $$ Solving: $$ f_1 = \frac{1}{2 \pi R C} $$ At this point:- The capacitive reactance equals the resistance.
- The impedance magnitude is equal parts resistive and capacitive, and the phase angle is about ($-45^\circ$).
Let's Do Some Maths
Thus far, the analysis of series circuits has been limited to a particular frequency. We will now examine the effect of frequency on the response of an RC series configuration such as that in [Fig. 1]. The magnitude of the source is fixed at 10 V, but the frequency range of analysis will extend from zero to 20 kHz.
Fig. 1: Determining the frequency response of a series RC circuit
Fig. 2: The frequency response of the individual elements of a series RC circuit.
$$X_C = { 1 \over 2 \pi f_1 C} = R$$
$$\bbox[10px,border:1px solid grey]{f_1 = { 1 \over 2 \pi R C}}$$
$$f_1 = { 1 \over 2 \pi (5kΩ)(0.01 \mu F)}=3183.1 Hz$$
Now for the details. The total impedance is determined by the following equation:
$$Z_T = R - j X_C$$
$$\bbox[10px,border:1px solid grey]{Z_T = Z_T \angle \theta_T\\
Z_T = \sqrt{ R^2 + X_C^2} \angle -tan^{-1}{X_C \over R}} \tag{1}$$
f = 100 Hz
$$X_C = { 1 \over 2 \pi f C} = { 1 \over 2 \pi (100Hz)(0.01 \mu F)} = 159.16 kΩ $$
$$Z_T = \sqrt{R^2 + X^2}\\
=\sqrt{(5kΩ)^2+(159.16kΩ)^2} = 159.24 kΩ$$
$$\theta_T = -tan^{-1}{X_C \over R} = -tan^{-1}{159.16k \over 5k}\\
=-tan^{-1} 31.83 = -88.2^\circ$$
$$ Z_T = 159.24 k Ω \angle 88.2^\circ$$
f = 1 kHz
$$X_C = { 1 \over 2 \pi f C} = { 1 \over 2 \pi (1000Hz)(0.01 \mu F)} = 15.92 kΩ $$
$$Z_T = \sqrt{R^2 + X^2}\\
=\sqrt{(5kΩ)^2+(15.92 kΩ)^2} = 16.69 kΩ$$
$$\theta_T = -tan^{-1}{X_C \over R} = -tan^{-1}{15.92k \over 5k}\\
=-tan^{-1} 3.18 = -72.54^\circ$$
$$ Z_T = 16.69 k Ω \angle -72.54^\circ$$
$$f = 5 kHz: Z_T = 5.93 k Ω \angle -32.48^\circ$$
$$f = 10 kHz: Z_T = 5.25 k Ω \angle -17.66^\circ$$
$$f = 15 kHz: Z_T = 5.11 k Ω \angle -11.98^\circ$$
$$f = 20 kHz: Z_T = 5.06 k Ω \angle -9.04^\circ$$
A plot of $Z_T$ versus frequency in [Fig. 3] completely supports our assumption based on the curves of [Fig. 2].
Fig. 3: The magnitude of the input impedance versus frequency for the circuit of Fig. 1.
Be the first to comment here!

Do you have any questions?