00:01
Suppose an incident wave is sent of the specified shape given by g z minus v1t downstream number 1.
00:14
It gives rise to a reflected wave given by hr z plus v1t and a transmitted wave given by gtz minus v2t.
00:28
By imposing the boundary conditions, find h .r and g .t.
00:40
The shape of incident wave is given by g .i.
00:46
Z.
00:46
Minus v1t.
00:49
This is the reflected wave, h .r.
00:51
Z plus vanty, and the transmitted wave is gt, z minus v2 t.
00:57
The imposing boundary conditions, which are fz, comma, t, is continued.
01:02
At z equals to 0 and f of 0 t is equals to f of 0 plus t and del f by del z at 0 negative is equal to del f by del z at 0 plus the derivative of f must also be continuous the transmitted wave is the sum of incident and reflected so applying the boundary conditions g .i.
01:36
0 minus v1 t plus hr 0 plus v1 t equals to g t 0 minus v2 t so the equation reduced to g i minus of v1 t plus hr into v1 t equals g t into minus of v2 t applying the boundary condition 2, that is finding the derivative, it will be del of g .i.
02:14
Minus v1t by del t minus del t.
02:20
H .r.
02:21
V1t by del t equals v1 by v2, del g t, v2, integrating on both the sides, the value of integration will be equals to gi minus v1t minus hr v1t equals v1 by v2 gt minus of v2t plus k, where k is the integration constant.
02:59
Adding the equations the first and the second one adding these two equations will give us the value...