Mixing
Simplifying equation
$begingroup$
I'm looking at answers to a question and I see these two lines of the equation. How is they're getting the -0.5095 on the bottom?
eqn
bilinear-form
asked Jan 10 at 14:14
controlled_controlled_
111
$endgroup$
add a comment |
$begingroup$
I'm looking at answers to a question and I see these two lines of the equation. How is they're getting the -0.5095 on the bottom?
eqn
bilinear-form
asked Jan 10 at 14:14
controlled_controlled_
111
$endgroup$
$begingroup$
They multiply both numerator and denominator by $frac 1{1.3249}=0.754773945$
$endgroup$
– lulu
Jan 10 at 14:17
add a comment |
$begingroup$
I'm looking at answers to a question and I see these two lines of the equation. How is they're getting the -0.5095 on the bottom?
eqn
bilinear-form
asked Jan 10 at 14:14
controlled_controlled_
111
$endgroup$
I'm looking at answers to a question and I see these two lines of the equation. How is they're getting the -0.5095 on the bottom?
eqn
bilinear-form
bilinear-form
asked Jan 10 at 14:14
controlled_controlled_
111
asked Jan 10 at 14:14
controlled_controlled_
111
asked Jan 10 at 14:14
controlled_controlled_
111
asked Jan 10 at 14:14
controlled_controlled_
111
asked Jan 10 at 14:14
controlled_controlled_
111
111
$begingroup$
They multiply both numerator and denominator by $frac 1{1.3249}=0.754773945$
$endgroup$
– lulu
Jan 10 at 14:17
add a comment |
$begingroup$
They multiply both numerator and denominator by $frac 1{1.3249}=0.754773945$
$endgroup$
– lulu
Jan 10 at 14:17
$begingroup$
They multiply both numerator and denominator by $frac 1{1.3249}=0.754773945$
$endgroup$
– lulu
Jan 10 at 14:17
$begingroup$
They multiply both numerator and denominator by $frac 1{1.3249}=0.754773945$
$endgroup$
– lulu
Jan 10 at 14:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$require{cancel}H(z)=frac{0.3249(z+1)}{1.3249z-0.6751}cdotfrac{color{red}{frac1{1.3249}}}{color{red}{frac1{1.3249}}}=frac{frac{{0.3249}}{color{red}{1.3249}}(z+1)}{cancel{frac{1.3249}{color{red}{1.3249}}}z-frac{0.6751}{color{red}{1.3249}}}=frac{0.2452(z+1)}{z-0.5095}$$
answered Jan 10 at 14:27
John GlennJohn Glenn
1,956424
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
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$begingroup$
$$require{cancel}H(z)=frac{0.3249(z+1)}{1.3249z-0.6751}cdotfrac{color{red}{frac1{1.3249}}}{color{red}{frac1{1.3249}}}=frac{frac{{0.3249}}{color{red}{1.3249}}(z+1)}{cancel{frac{1.3249}{color{red}{1.3249}}}z-frac{0.6751}{color{red}{1.3249}}}=frac{0.2452(z+1)}{z-0.5095}$$
answered Jan 10 at 14:27
John GlennJohn Glenn
1,956424
$endgroup$
add a comment |
$begingroup$
$$require{cancel}H(z)=frac{0.3249(z+1)}{1.3249z-0.6751}cdotfrac{color{red}{frac1{1.3249}}}{color{red}{frac1{1.3249}}}=frac{frac{{0.3249}}{color{red}{1.3249}}(z+1)}{cancel{frac{1.3249}{color{red}{1.3249}}}z-frac{0.6751}{color{red}{1.3249}}}=frac{0.2452(z+1)}{z-0.5095}$$
answered Jan 10 at 14:27
John GlennJohn Glenn
1,956424
$endgroup$
add a comment |
$begingroup$
$$require{cancel}H(z)=frac{0.3249(z+1)}{1.3249z-0.6751}cdotfrac{color{red}{frac1{1.3249}}}{color{red}{frac1{1.3249}}}=frac{frac{{0.3249}}{color{red}{1.3249}}(z+1)}{cancel{frac{1.3249}{color{red}{1.3249}}}z-frac{0.6751}{color{red}{1.3249}}}=frac{0.2452(z+1)}{z-0.5095}$$
answered Jan 10 at 14:27
John GlennJohn Glenn
1,956424
$endgroup$
$$require{cancel}H(z)=frac{0.3249(z+1)}{1.3249z-0.6751}cdotfrac{color{red}{frac1{1.3249}}}{color{red}{frac1{1.3249}}}=frac{frac{{0.3249}}{color{red}{1.3249}}(z+1)}{cancel{frac{1.3249}{color{red}{1.3249}}}z-frac{0.6751}{color{red}{1.3249}}}=frac{0.2452(z+1)}{z-0.5095}$$
answered Jan 10 at 14:27
John GlennJohn Glenn
1,956424
answered Jan 10 at 14:27
John GlennJohn Glenn
1,956424
answered Jan 10 at 14:27
John GlennJohn Glenn
1,956424
answered Jan 10 at 14:27
John GlennJohn Glenn
1,956424
1,956424
add a comment |
add a comment |
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$begingroup$
They multiply both numerator and denominator by $frac 1{1.3249}=0.754773945$
$endgroup$
– lulu
Jan 10 at 14:17