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$$
\text { Assume that the numerical values for the signal voltage shown in Fig. are } V_{m}=50 \mathrm{mV} \text { and } t_{1}=1 \mathrm{s}.
$$

$$
\begin{array}{l}{\text { This signal voltage is applied to integrating-amplifier circuit as input. }} \\ {\text { The circuit parameters of the amplifier are } R_{s}=100 \mathrm{k} \Omega, C_{f}=0.1 \mu \mathrm{F} \text { , }} \\ {\text { and } V_{c c}=6 \mathrm{V} \text { . The initial voltage on the capacitor is zero. }} \\ {\text { a) Calculate } V_{o}(t) .} \\ {\text { b) Plot } V_{o(t)} \text { versus } t}\end{array}
$$

$$
t_{1}=1 \sec ، V_{m}=50 \mathrm{mV}
$$

$$
V_{0}(t)=\frac{-1}{R_{s} C_{f}} \int_{0}^{t} V_{s} d t
$$

\( V_{0}(t)=\frac{-1}{100 * 10^{3} * 0.1 *10^{-6}}\int_{0}^{t}\)  

 

                                                        

(1) $$
0 \leq t \leq 1 \text { sec }
$$                                        $$
V_{s}=50 m v
$$

$$
V_{0}(t)=-100+\int_{0}^{t} 50 \times 10^{-3} d t
$$

$$
V_{0}(t)=-5 t \Rightarrow 0 \leq t \leq 1 \sec
$$

(2) $$
1 \leq t \leq 2 \text { sec }
$$                        $$
V_{s}=-50 \mathrm{mv}
$$

$$
V_{0}(t)=\int^{t}_{t_0}V_s dt*\frac{-1}{R_sC_f}+V(t_0)
$$

$$
V_{0}(t)=\int_{1}^{t}-50 * 10^{-3} *-100+-5)
$$

$$
V_0(t)=5(t-1)-5
$$

$$
V_{0}(t)=\left\{\begin{array}{cc}{-5 t} & {0 \leq t \leq 1} \\ {5(t-1)-5} & {1 \leq t \leq 2}\end{array}\right.
$$

$$
V_{0}=-5 t
$$

$$
t=0
\quad \longrightarrow V_{0}=0 \quad(0,0)
$$

$$
t=1
 \quad \longrightarrow V_{0}=-5 \quad(1,-5)
$$

$$
V_{0}=5(t-1)-5
$$

$$
t=1
$$                                $$
V_{0}=-5(1,-5)
$$

$$
t=2
$$                                $$
V_{0}=0(2,0)
$$

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