diff options
Diffstat (limited to 'source/know/concept/convolution-theorem')
| -rw-r--r-- | source/know/concept/convolution-theorem/index.md | 18 |
1 files changed, 9 insertions, 9 deletions
diff --git a/source/know/concept/convolution-theorem/index.md b/source/know/concept/convolution-theorem/index.md index 510417a..d10d85d 100644 --- a/source/know/concept/convolution-theorem/index.md +++ b/source/know/concept/convolution-theorem/index.md @@ -36,9 +36,9 @@ rearrange the integrals: $$\begin{aligned} \hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} - &= B \int_{-\infty}^\infty \tilde{f}(k) \Big( A \int_{-\infty}^\infty g(x') \exp(i s k x') \dd{x'} \Big) \exp(-i s k x) \dd{k} + &= B \int_{-\infty}^\infty \tilde{f}(k) \Big( A \int_{-\infty}^\infty g(x') \: e^{i s k x'} \dd{x'} \Big) e^{-i s k x} \dd{k} \\ - &= A \int_{-\infty}^\infty g(x') \Big( B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k (x - x')) \dd{k} \Big) \dd{x'} + &= A \int_{-\infty}^\infty g(x') \Big( B \int_{-\infty}^\infty \tilde{f}(k) \: e^{-i s k (x - x')} \dd{k} \Big) \dd{x'} \\ &= A \int_{-\infty}^\infty g(x') \: f(x - x') \dd{x'} = A \cdot (f * g)(x) @@ -49,9 +49,9 @@ this time starting from a product in the $$x$$-domain: $$\begin{aligned} \hat{\mathcal{F}}\{f(x) \: g(x)\} - &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \exp(- i s x k') \dd{k'} \Big) \exp(i s k x) \dd{x} + &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \: e^{-i s x k'} \dd{k'} \Big) e^{i s k x} \dd{x} \\ - &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \exp(i s x (k - k')) \dd{x} \Big) \dd{k'} + &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \: e^{i s x (k - k')} \dd{x} \Big) \dd{k'} \\ &= B \int_{-\infty}^\infty \tilde{g}(k') \: \tilde{f}(k - k') \dd{k'} = B \cdot (\tilde{f} * \tilde{g})(k) @@ -86,20 +86,20 @@ because we set both $$f(t)$$ and $$g(t)$$ to zero for $$t < 0$$: $$\begin{aligned} \hat{\mathcal{L}}\{(f * g)(t)\} - &= \int_0^\infty \Big( \int_0^\infty g(t') f(t - t') \dd{t'} \Big) \exp(- s t) \dd{t} + &= \int_0^\infty \Big( \int_0^\infty g(t') \: f(t - t') \dd{t'} \Big) e^{-s t} \dd{t} \\ - &= \int_0^\infty \Big( \int_0^\infty f(t - t') \exp(- s t) \dd{t} \Big) g(t') \dd{t'} + &= \int_0^\infty \Big( \int_0^\infty f(t - t') \: e^{-s t} \dd{t} \Big) g(t') \dd{t'} \end{aligned}$$ Then we define a new integration variable $$\tau = t - t'$$, yielding: $$\begin{aligned} \hat{\mathcal{L}}\{(f * g)(t)\} - &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s (\tau + t')) \dd{\tau} \Big) g(t') \dd{t'} + &= \int_0^\infty \Big( \int_0^\infty f(\tau) \: e^{-s (\tau + t')} \dd{\tau} \Big) g(t') \dd{t'} \\ - &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s \tau) \dd{\tau} \Big) g(t') \exp(- s t') \dd{t'} + &= \int_0^\infty \Big( \int_0^\infty f(\tau) \: e^{-s \tau} \dd{\tau} \Big) g(t') \: e^{-s t'} \dd{t'} \\ - &= \int_0^\infty \tilde{f}(s) \: g(t') \exp(- s t') \dd{t'} + &= \int_0^\infty \tilde{f}(s) \: g(t') \: e^{-s t'} \dd{t'} = \tilde{f}(s) \: \tilde{g}(s) \end{aligned}$$ {% include proof/end.html id="proof-laplace" %} |