(You have to make manuel calculations for his part) 1. Assume that Vin is a sine wave with the amplitude of 400mVp-p,Rin=1kOmega ,Rf=100kOmega ,Vcc=15V in Fig 2.4. Find the output peak to peak amplitude value. 2. For inverting integrator circuit in Fig2.6 investigate why the feedback resistor (Rf) is added to the circuit.

1. To find the output peak-to-peak amplitude value, we can use the formula:<br /><br />\[ V_{out\_pp} = V_{in\_pp} \times \frac{Rf}{Rin} \]<br /><br />Given:<br />- \( V_{in\_pp} = 400mV \)<br />- \( Rin = 1k\Omega \)<br />- \( Rf = 100k\Omega \)<br /><br />Substituting the values:<br /><br />\[ V_{out\_pp} = 400mV \times \frac{100k\Omega}{1k\Omega} \]<br />\[ V_{out\_pp} = 400mV \times 100 \]<br />\[ V_{out\_pp} = 40V \]<br /><br />Therefore, the output peak-to-peak amplitude value is 40V.<br /><br />2. In an inverting integrator circuit, the feedback resistor \( Rf \) is added to the circuit to control the rate of integration. The feedback resistor \( Rf \) determines the time constant of the integrator circuit, which is given by \( Rf \times C \), where \( C \) is the capacitance of the capacitor in the circuit.<br /><br />By adjusting the value of \( Rf \), we can control the rate at which the voltage across the capacitor increases or decreases. This allows us to adjust the frequency response of the circuit and shape the output waveform as desired. The feedback resistor \( Rf \) helps to stabilize the circuit and improve its overall performance.