Max Woesner Find F [ c o s ( w 0 t ) g ( t ) ] {\displaystyle {\mathcal {F}}[cos(w_{0}t)g(t)]\!} Recall w 0 = 2 π f 0 {\displaystyle w_{0}=2\pi f_{0}\!} , so F [ c o s ( w 0 t ) g ( t ) ] = F [ c o s ( 2 π f 0 t ) g ( t ) ] = ∫ − ∞ ∞ c o s ( 2 π f 0 t ) g ( t ) e − j 2 π f t d t {\displaystyle {\mathcal {F}}[cos(w_{0}t)g(t)]={\mathcal {F}}[cos(2\pi f_{0}t)g(t)]=\int _{-\infty }^{\infty }cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt\!} Also recall c o s ( θ ) = 1 2 ( e j θ + e − j θ ) {\displaystyle cos(\theta )={\frac {1}{2}}(e^{j\theta }+e^{-j\theta })\!} ,so ∫ − ∞ ∞ c o s ( 2 π f 0 t ) g ( t ) e − j 2 π f t d t = ∫ − ∞ ∞ 1 2 [ e j 2 π f 0 t + e − j 2 π f 0 t ] g ( t ) e − j 2 π f t d t {\displaystyle \int _{-\infty }^{\infty }cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt=\int _{-\infty }^{\infty }{\frac {1}{2}}[e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}]g(t)e^{-j2\pi ft}dt\!} Now ∫ − ∞ ∞ 1 2 [ e j 2 π f 0 t + e − j 2 π f 0 t ] g ( t ) e − j 2 π f t d t = 1 2 ∫ − ∞ ∞ e − j 2 π ( f − f 0 ) t g ( t ) d t + 1 2 ∫ − ∞ ∞ e − j 2 π ( f + f 0 ) t g ( t ) d t = 1 2 G ( f − f 0 ) + 1 2 G ( f + f 0 ) {\displaystyle \int _{-\infty }^{\infty }{\frac {1}{2}}[e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}]g(t)e^{-j2\pi ft}dt={\frac {1}{2}}\int _{-\infty }^{\infty }e^{-j2\pi (f-f_{0})t}g(t)dt+{\frac {1}{2}}\int _{-\infty }^{\infty }e^{-j2\pi (f+f_{0})t}g(t)dt={\frac {1}{2}}G(f-f_{0})+{\frac {1}{2}}G(f+f_{0})\!} So F [ c o s ( w 0 t ) g ( t ) ] = 1 2 [ G ( f − f 0 ) + G ( f + f 0 ) ] {\displaystyle {\mathcal {F}}[cos(w_{0}t)g(t)]={\frac {1}{2}}[G(f-f_{0})+G(f+f_{0})]\!}
Nick Christman 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})}