Frequency response simple equation
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I'm looking at answers to calculating the frequency response, but i don't understand one line. what are they doing to go from the first line to the second line?
functions
asked Jan 10 at 15:52
controlled_controlled_
111
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add a comment |
$begingroup$
I'm looking at answers to calculating the frequency response, but i don't understand one line. what are they doing to go from the first line to the second line?
functions
asked Jan 10 at 15:52
controlled_controlled_
111
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1
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What is blocking you ? $e^{-ja/2}, e^{ja/2} = 1$ for example
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– Damien
Jan 10 at 15:57
add a comment |
$begingroup$
I'm looking at answers to calculating the frequency response, but i don't understand one line. what are they doing to go from the first line to the second line?
functions
asked Jan 10 at 15:52
controlled_controlled_
111
$endgroup$
I'm looking at answers to calculating the frequency response, but i don't understand one line. what are they doing to go from the first line to the second line?
functions
functions
asked Jan 10 at 15:52
controlled_controlled_
111
asked Jan 10 at 15:52
controlled_controlled_
111
asked Jan 10 at 15:52
controlled_controlled_
111
asked Jan 10 at 15:52
controlled_controlled_
111
asked Jan 10 at 15:52
controlled_controlled_
111
111
1
$begingroup$
What is blocking you ? $e^{-ja/2}, e^{ja/2} = 1$ for example
$endgroup$
– Damien
Jan 10 at 15:57
add a comment |
1
$begingroup$
What is blocking you ? $e^{-ja/2}, e^{ja/2} = 1$ for example
$endgroup$
– Damien
Jan 10 at 15:57
1
1
$begingroup$
What is blocking you ? $e^{-ja/2}, e^{ja/2} = 1$ for example
$endgroup$
– Damien
Jan 10 at 15:57
$begingroup$
What is blocking you ? $e^{-ja/2}, e^{ja/2} = 1$ for example
$endgroup$
– Damien
Jan 10 at 15:57
add a comment |
1 Answer
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They are drawing out a factor of $e^{-jomega/2}$ from the expression inside the brackets,$$H(omega)=0.5(1+e^{-jomega})cdotfrac{e^{-jomega/2}}{e^{-jomega/2}}=0.5e^{-jomega/2}Big(e^{jomega/2}+e^{-jomega/2}Big)$$
answered Jan 10 at 15:57
Shubham JohriShubham Johri
5,560818
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$begingroup$
They are drawing out a factor of $e^{-jomega/2}$ from the expression inside the brackets,$$H(omega)=0.5(1+e^{-jomega})cdotfrac{e^{-jomega/2}}{e^{-jomega/2}}=0.5e^{-jomega/2}Big(e^{jomega/2}+e^{-jomega/2}Big)$$
answered Jan 10 at 15:57
Shubham JohriShubham Johri
5,560818
$endgroup$
add a comment |
$begingroup$
They are drawing out a factor of $e^{-jomega/2}$ from the expression inside the brackets,$$H(omega)=0.5(1+e^{-jomega})cdotfrac{e^{-jomega/2}}{e^{-jomega/2}}=0.5e^{-jomega/2}Big(e^{jomega/2}+e^{-jomega/2}Big)$$
answered Jan 10 at 15:57
Shubham JohriShubham Johri
5,560818
$endgroup$
add a comment |
$begingroup$
They are drawing out a factor of $e^{-jomega/2}$ from the expression inside the brackets,$$H(omega)=0.5(1+e^{-jomega})cdotfrac{e^{-jomega/2}}{e^{-jomega/2}}=0.5e^{-jomega/2}Big(e^{jomega/2}+e^{-jomega/2}Big)$$
answered Jan 10 at 15:57
Shubham JohriShubham Johri
5,560818
$endgroup$
They are drawing out a factor of $e^{-jomega/2}$ from the expression inside the brackets,$$H(omega)=0.5(1+e^{-jomega})cdotfrac{e^{-jomega/2}}{e^{-jomega/2}}=0.5e^{-jomega/2}Big(e^{jomega/2}+e^{-jomega/2}Big)$$
answered Jan 10 at 15:57
Shubham JohriShubham Johri
5,560818
answered Jan 10 at 15:57
Shubham JohriShubham Johri
5,560818
answered Jan 10 at 15:57
Shubham JohriShubham Johri
5,560818
answered Jan 10 at 15:57
Shubham JohriShubham Johri
5,560818
5,560818
add a comment |
add a comment |
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$begingroup$
What is blocking you ? $e^{-ja/2}, e^{ja/2} = 1$ for example
$endgroup$
– Damien
Jan 10 at 15:57