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## Q. 7.3

Consider the RC circuit shown in Fig. 7.15. Calculate the noise spectrum and the total noise power in $V_{out}$.

## Verified Solution

We follow the four steps described in Section 7.1.5. The noise spectrum of R is given by $S_v( f ) = 4kT R.$ Next, modeling the noise of R by a series voltage source $V_R$, we compute the transfer function from $V_R to V_{out}$:

$\frac{V_{out}}{V_R}(s) =\frac{1}{RCs + 1}$                                 (7.17)

From the theorem in Section 7.1.1, we have

$S_{out}( f ) = S_v( f )\left|\frac{V_{out}}{V_R}(jω)\right|^2$                   (7.18)

$= 4kT R \frac{1}{4π^2R^2C^2 f ^2 + 1}$                       (7.19)

Thus, the white noise spectrum of the resistor is shaped by a low-pass characteristic (Fig. 7.16). To calculate the total noise power at the output, we write

$P_{n,out} =\int_{0}^{\infty }{\frac{4kT R}{4π^2R^2C^2 f^2 + 1}d f}$                                (7.20)

Note that the integration must be with respect to f rather than ω (why?). Since

$\int{\frac{dx}{x^2 + 1}} = \tan^{−1} x$                              (7.21)

the integral reduces to

\begin{aligned} P_{n, \text { out }} &=\left.\frac{2 k T}{\pi C} \tan ^{-1} u\right|_{u=0} ^{u=\infty} (7.22) \\ &=\frac{k T}{C} \end{aligned} (7.23)

Note that the unit of k T / C is $V^2.$ We may also consider $\sqrt{k T / C}$ as the total rms noise voltage measured at the output. For example, with a 1-pF capacitor, the total noise voltage is equal to $64.3 \mu V_ {rms }$ at $T=300 \mathrm{~K}.$

Equation (7.23) implies that the total noise at the output of the circuit shown in Fig. 7.15 is independent of the value of R. Intuitively, this is because for larger values of R, the associated noise per unit bandwidth increases while the overall bandwidth of the circuit decreases. The fact that kT / C noise can be decreased only by increasing C (if T is fixed) introduces many difficulties in the design of analog circuits (Chapter 13).