Fourier Transform Properties
Find F [ 10 t g ( t ) e j 2 π f t 0 ] {\displaystyle {\mathcal {F}}[10^{t}g(t)e^{j2\pi ft_{0}}]}
To begin, we know that
F [ 10 t g ( t ) e j 2 π f t 0 ] = ∫ − ∞ ∞ 10 t g ( t ) e j 2 π f t 0 e − j 2 π f t d t = ∫ − ∞ ∞ 10 t g ( t ) e j 2 π f ( t 0 − t ) d t {\displaystyle {\mathcal {F}}[10^{t}g(t)e^{j2\pi ft_{0}}]=\int _{-\infty }^{\infty }10^{t}g(t)e^{j2\pi ft_{0}}e^{-j2\pi ft}\,dt=\int _{-\infty }^{\infty }10^{t}g(t)e^{j2\pi f(t_{0}-t)}\,dt}
But recall that e j 2 π f ( t 0 − t ) ≡ δ ( t 0 − t ) or δ ( t − t 0 ) {\displaystyle e^{j2\pi f(t_{0}-t)}\equiv \delta (t_{0}-t){\mbox{ or }}\delta (t-t_{0})}
Because of this definition, our problem has now been simplified significantly:
F [ 10 t g ( t ) e j 2 π f t 0 ] = ∫ − ∞ ∞ 10 t g ( t ) δ ( t − t 0 ) d t = 10 t 0 g ( t 0 ) {\displaystyle {\mathcal {F}}[10^{t}g(t)e^{j2\pi ft_{0}}]=\int _{-\infty }^{\infty }10^{t}g(t)\delta (t-t_{0})\,dt=10^{t_{0}}g(t_{0})}
Therefore,
F [ 10 t g ( t ) e j 2 π f t 0 ] = 10 t 0 g ( t 0 ) {\displaystyle {\mathcal {F}}[10^{t}g(t)e^{j2\pi ft_{0}}]=10^{t_{0}}g(t_{0})}
![{\displaystyle {\mathcal {F}}[10^{t}g(t)e^{j2\pi ft_{0}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/970f0d2e311f207b67e07a1b4d1750bea8f28b72)
![{\displaystyle {\mathcal {F}}[10^{t}g(t)e^{j2\pi ft_{0}}]=\int _{-\infty }^{\infty }10^{t}g(t)e^{j2\pi ft_{0}}e^{-j2\pi ft}\,dt=\int _{-\infty }^{\infty }10^{t}g(t)e^{j2\pi f(t_{0}-t)}\,dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f457100a5baa4e23ec88542e5d21777e9856c4c4)
![{\displaystyle {\mathcal {F}}[10^{t}g(t)e^{j2\pi ft_{0}}]=\int _{-\infty }^{\infty }10^{t}g(t)\delta (t-t_{0})\,dt=10^{t_{0}}g(t_{0})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/61f146313bd61889a8c81a215a477fe32a724a1a)
![{\displaystyle {\mathcal {F}}[10^{t}g(t)e^{j2\pi ft_{0}}]=10^{t_{0}}g(t_{0})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/077080dd33a4b881b49124d1f596a5df4bacd59b)