Given data:

Refer to Figure 6.91 in the textbook.

Formula used:

Write the expression to calculate the straight line equation for two points (x1,y1) and (x2,y2).

(y−y1)=y2−y1x2−x1(x−x1) (1)

Refer to Figure 6.91(a) in the textbook.

From the given voltage graph, substitute t for x and vi(t) for y in equation (1).

(vi(t)−y1)=y2−y1x2−x1(t−x1) (2)

Write the expression to calculate the output voltage of an op amp differentiator.

vo(t)=−RCdvi(t)dt (3)

Here,

R is the value of the resistor,

C is the value of the capacitor, and

vi is the input voltage.

Calculation:

The given circuit is redrawn as Figure 1.

Substitute 20 kΩ for R and 0.01 μF for C in equation (3) to find vo(t).

vo(t)=−(20 kΩ)(0.01 μF)dvi(t)dt=−(20×103)(0.01×10−6)dvi(t)dt Ω⋅F {1 k=103, 1 μ=10−6}

vo(t)=−(2×10−4) dvi(t)dt Ω⋅F (4)

The given input voltage waveform is redrawn as Figure 2.

Refer to Figure 2, split up the time period as three divisions such as 0 ms<t<1 ms, 1 ms<t<3 ms and 3 ms<t<4 ms to find the respective voltage value.

Case (i): 0 ms<t<1 ms

The two points (x1,y1) and (x2,y2) are (0 ms,0 V) and (1 ms,10 V).

Substitute 0 ms for x1, 0 V for y1, 1 ms for x2 and 10 V for y2 in equation (2) to find vi(t).

(vi(t)−0 V)=10 V−0 V1 ms−0 ms(t−0 ms)=10 V1 ms(t)=101×10−3(t) V {∵1 m=10−3}=10000 t V

Simplify the equation to find vi(t).

vi(t)=10000 t V+0 V=10000 t V

Case (ii): 1 ms<t<3 ms

The two points (x1,y1) and (x2,y2) are (1 ms,10 V) and (3 ms,−10 V).

Substitute 1 ms for x1, 10 V for y1, 3 ms for x2 and −10 V for y2 in equation (2) to find vi(t).

(vi(t)−10 V)=−10 V−10 V3 ms−1 ms(t−1 ms)=−20 V2 ms(t−1 ms)=−20(2×10−3)(t−(1×10−3)) V {∵1 m=10−3}=−10000t V+0.02 V

Simplify the equation to find vi(t).

vi(t)=−10000t V+0.02 V+10 V=−10000t V+10.02 V=(−10000t+10.02) V

Case (iii): 3 ms<t<4 ms

The two points (x1,y1) and (x2,y2) are (3 ms,−10 V) and (4 ms,0 V).

Substitute 3 ms for x1, −10 V for y1, 4 ms for x2 and 0 V for y2 in equation (2) to find vi(t).

(vi(t)−(−10 V))=0 V−(−10 V)4 ms−3 ms(t−3 ms)(vi(t)+10 V)=10 V1 ms(t−3 ms)=101×10−3(t−(3×10−3)) V {∵1 m=10−3}=10000t V−0.03 V

Simplify the equation to find vi(t).

vi(t)=10000t V−0.03 V−10 V=10000t V−10.03 V=(10000t−10.03) V

Therefore, the input voltage function vi(t) of the signal in Figure 2 is,

vi(t)={10000 t V, 0 ms<t<1 ms(−10000t+10.02) V, 1 ms<t<3 ms(10000t−10.03) V, 3 ms<t<4 ms

For 0 ms<t<1 ms:

Substitute 10000 t V for vi(t) in equation (4) to find vo(t).

vo(t)=−(2×10−4) d(10000 t V)dt Ω⋅F=−(2×10−4)(10000) ddt(t) Ω⋅F⋅V=−2 (1) Ω⋅F⋅Vs=−2 F⋅V(sΩ)

Simplify the equation to find vo(t).

vo(t)=−2 F⋅VF {∵1 F=1s1Ω}=−2 V

For 1 ms<t<3 ms:

Substitute (−10000t+10.02) V for vi(t) in equation (4) to find vo(t).

vo(t)=−(2×10−4) d((−10000t+10.02) V)dt Ω⋅F=−(2×10−4) ddt(−10000t+10.02) Ω⋅F⋅V=−(2×10−4) (−10000(1)+0) Ω⋅F⋅Vs=−(2×10−4)(−10000) F⋅V(sΩ)

Simplify the equation to find vo(t).

vo(t)=2 F⋅VF {∵1 F=1s1Ω}=2 V

For 3 ms<t<4 ms:

Substitute (10000t−10.03) V for vi(t) in equation (4) to find vo(t).

vo(t)=−(2×10−4) d((10000t−10.03) V)dt Ω⋅F=−(2×10−4) ddt(10000t−10.03) Ω⋅F⋅V=−(2×10−4) (10000(1)−0) Ω⋅F⋅Vs=−(2×10−4)(10000) F⋅V(sΩ)

Simplify the equation to find vo(t).

vo(t)=−2 F⋅VF {∵1 F=1s1Ω}=−2 V

Therefore, the output voltage function vo(t) is expressed as,

vo(t)={−2 V, 0 ms<t<1 ms2 V, 1 ms<t<3 ms−2 V, 3 ms<t<4 ms

The output voltage waveform is drawn as Figure 3.

Conclusion:

Thus, the output voltage waveform of an op amp differentiator vo(t) is sketched.