Fourier transformation with two signals
So I have two signals
$$m(t) = 336cdotrm{sinc}(24t)$$
$$x(t) = m(t)cos(400pi t)$$
How do I calculate fourier transformation for $x(t)$?
I already calculated:
$$M(f) = 14cdotrm{rect}(frac{f}{24})$$
and fourier transformation for $cos(400 pi t)$, lets just call this $y(t)$
$$Y(f) = frac{1}{2}(delta(f - 200) + delta(f + 200))$$
$$X(f) = 14rm{rect}(frac{f}{24}) cdot frac{1}{2}(delta(f - 200) + delta(f + 200))$$
which is wrong answer. What am I doing wrong?
fourier-transform
edited Nov 18 at 17:14
Trebor
58112
asked Nov 18 at 16:30
Math101
31
add a comment |
So I have two signals
$$m(t) = 336cdotrm{sinc}(24t)$$
$$x(t) = m(t)cos(400pi t)$$
How do I calculate fourier transformation for $x(t)$?
I already calculated:
$$M(f) = 14cdotrm{rect}(frac{f}{24})$$
and fourier transformation for $cos(400 pi t)$, lets just call this $y(t)$
$$Y(f) = frac{1}{2}(delta(f - 200) + delta(f + 200))$$
$$X(f) = 14rm{rect}(frac{f}{24}) cdot frac{1}{2}(delta(f - 200) + delta(f + 200))$$
which is wrong answer. What am I doing wrong?
fourier-transform
edited Nov 18 at 17:14
Trebor
58112
asked Nov 18 at 16:30
Math101
31
add a comment |
So I have two signals
$$m(t) = 336cdotrm{sinc}(24t)$$
$$x(t) = m(t)cos(400pi t)$$
How do I calculate fourier transformation for $x(t)$?
I already calculated:
$$M(f) = 14cdotrm{rect}(frac{f}{24})$$
and fourier transformation for $cos(400 pi t)$, lets just call this $y(t)$
$$Y(f) = frac{1}{2}(delta(f - 200) + delta(f + 200))$$
$$X(f) = 14rm{rect}(frac{f}{24}) cdot frac{1}{2}(delta(f - 200) + delta(f + 200))$$
which is wrong answer. What am I doing wrong?
fourier-transform
edited Nov 18 at 17:14
Trebor
58112
asked Nov 18 at 16:30
Math101
31
So I have two signals
$$m(t) = 336cdotrm{sinc}(24t)$$
$$x(t) = m(t)cos(400pi t)$$
How do I calculate fourier transformation for $x(t)$?
I already calculated:
$$M(f) = 14cdotrm{rect}(frac{f}{24})$$
and fourier transformation for $cos(400 pi t)$, lets just call this $y(t)$
$$Y(f) = frac{1}{2}(delta(f - 200) + delta(f + 200))$$
$$X(f) = 14rm{rect}(frac{f}{24}) cdot frac{1}{2}(delta(f - 200) + delta(f + 200))$$
which is wrong answer. What am I doing wrong?
fourier-transform
fourier-transform
edited Nov 18 at 17:14
Trebor
58112
asked Nov 18 at 16:30
Math101
31
edited Nov 18 at 17:14
Trebor
58112
asked Nov 18 at 16:30
Math101
31
edited Nov 18 at 17:14
Trebor
58112
edited Nov 18 at 17:14
Trebor
58112
edited Nov 18 at 17:14
Trebor
58112
58112
asked Nov 18 at 16:30
Math101
31
asked Nov 18 at 16:30
Math101
31
asked Nov 18 at 16:30
Math101
31
31
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Look up the convolution theorem for the Fourier Transform
$$begin{align*}X(f)&=mathscr{F}left{x(t)right}\
\
&= mathscr{F}left{m(t)cdot y(t)right}\
\
&=mathscr{F}left{m(t)right}*mathscr{F}left{ y(t)right}\
\
X(f) &= M(f) * Y(f)\
end{align*}$$
Where '$cdot$' denotes multiplication and '$*$' denotes convolution. Your answer just appears to have multiplication, given your notation.
answered Nov 18 at 17:07
Andy Walls
1,539128
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Look up the convolution theorem for the Fourier Transform
$$begin{align*}X(f)&=mathscr{F}left{x(t)right}\
\
&= mathscr{F}left{m(t)cdot y(t)right}\
\
&=mathscr{F}left{m(t)right}*mathscr{F}left{ y(t)right}\
\
X(f) &= M(f) * Y(f)\
end{align*}$$
Where '$cdot$' denotes multiplication and '$*$' denotes convolution. Your answer just appears to have multiplication, given your notation.
answered Nov 18 at 17:07
Andy Walls
1,539128
add a comment |
Look up the convolution theorem for the Fourier Transform
$$begin{align*}X(f)&=mathscr{F}left{x(t)right}\
\
&= mathscr{F}left{m(t)cdot y(t)right}\
\
&=mathscr{F}left{m(t)right}*mathscr{F}left{ y(t)right}\
\
X(f) &= M(f) * Y(f)\
end{align*}$$
Where '$cdot$' denotes multiplication and '$*$' denotes convolution. Your answer just appears to have multiplication, given your notation.
answered Nov 18 at 17:07
Andy Walls
1,539128
add a comment |
Look up the convolution theorem for the Fourier Transform
$$begin{align*}X(f)&=mathscr{F}left{x(t)right}\
\
&= mathscr{F}left{m(t)cdot y(t)right}\
\
&=mathscr{F}left{m(t)right}*mathscr{F}left{ y(t)right}\
\
X(f) &= M(f) * Y(f)\
end{align*}$$
Where '$cdot$' denotes multiplication and '$*$' denotes convolution. Your answer just appears to have multiplication, given your notation.
answered Nov 18 at 17:07
Andy Walls
1,539128
Look up the convolution theorem for the Fourier Transform
$$begin{align*}X(f)&=mathscr{F}left{x(t)right}\
\
&= mathscr{F}left{m(t)cdot y(t)right}\
\
&=mathscr{F}left{m(t)right}*mathscr{F}left{ y(t)right}\
\
X(f) &= M(f) * Y(f)\
end{align*}$$
Where '$cdot$' denotes multiplication and '$*$' denotes convolution. Your answer just appears to have multiplication, given your notation.
answered Nov 18 at 17:07
Andy Walls
1,539128
answered Nov 18 at 17:07
Andy Walls
1,539128
answered Nov 18 at 17:07
Andy Walls
1,539128
answered Nov 18 at 17:07
Andy Walls
1,539128
1,539128
add a comment |
add a comment |
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