Answer:
1st Harmonic:
![v(t) = 50\cos(2000\pi t)](https://img.qammunity.org/2021/formulas/engineering/middle-school/6jscuzvosu1kasc9cgu2r96u5ayhmu0ogg.png)
3rd Harmonic:
![v(t) = 9\cos(6000\pi t)](https://img.qammunity.org/2021/formulas/engineering/middle-school/pdlrvx9q8iau3y4h1t1570k4o85s3pr9ai.png)
5th Harmonic:
![v(t) = 6\cos(10000\pi t)](https://img.qammunity.org/2021/formulas/engineering/middle-school/es4kcbed132wmqacl0ptbqkyo2gt7l2tkv.png)
7th Harmonic:
![v(t) = 2\cos(14000\pi t)](https://img.qammunity.org/2021/formulas/engineering/middle-school/d0emvy0hkwxk3z3wvcelih4wpq5xoef8ki.png)
Step-by-step explanation:
The general form to represent a complex sinusoidal waveform is given by
![v(t) = A\cos(2\pi f t + \phi)](https://img.qammunity.org/2021/formulas/engineering/middle-school/lxkjc4d1bw134noem834pijf089dfeeiyl.png)
Where A is the amplitude in volts of the sinusoidal waveform
Where f is the frequency in cycles per second (Hz) of the sinusoidal waveform
Where
is the phase angle in radians of the sinusoidal waveform.
1st Harmonic:
We have A = 50, f = 1000 and φ = 0
![v(t) = 50\cos(2\pi 1000 t + 0) \\\\v(t) = 50\cos(2000\pi t)](https://img.qammunity.org/2021/formulas/engineering/middle-school/495c94l4qz7e3nxdv74duzpz2iwj83jxd7.png)
3rd Harmonic:
We have A = 9, f = 3000 and φ = 0
![v(t) = 9\cos(2\pi 3000 t + 0) \\\\v(t) = 9\cos(6000\pi t)](https://img.qammunity.org/2021/formulas/engineering/middle-school/r212ht3srb49w81zivcojxwkvu9b6qi49v.png)
5th Harmonic:
We have A = 6, f = 5000 and φ = 0
![v(t) = 6\cos(2\pi 5000 t + 0) \\\\v(t) = 6\cos(10000\pi t)](https://img.qammunity.org/2021/formulas/engineering/middle-school/eowlb6ouv6vgk6woat4wgo5gorfpqfiad6.png)
7th Harmonic:
We have A = 2, f = 7000 and φ = 0
![v(t) = 2\cos(2\pi 7000 t + 0) \\\\v(t) = 2\cos(14000\pi t)](https://img.qammunity.org/2021/formulas/engineering/middle-school/gtjorceoy2taa41la1je6gngn9ui1w80zx.png)
Note: The even-numbered harmonics have 0 amplitude that is why they are not shown here.