怎么在用于BASYS3项目的VHDL中创建3秒锁存器

我使用VHDL在FPGA上创建了一个音箱。
它有3个按钮和3个选择开关。
当任何选择器开关为高“1”时，与该开关相连的音调会在其为高电平时播放。
当它变低时，音调停止。
我希望我的按钮在切换为高'1'时，在按钮释放后保持高电平3秒，然后转到零。
有人称这是一次性，其中一个循环锁定一点。
我试图使用等待声明，但它不起作用。
只要按住按钮，音调就会播放，但在释放时会立即停止。
这是我的代码的一部分。
任何帮助表示赞赏。
注意：周期是一个常数（即'恒定周期：时间：= 1000毫秒;'）
D1：如果btn_a ='1'则进程开始，然后tempdata等待3 *周期;
elsif btn_b ='1'然后tempdata等待3 *期;
elsif btn_c ='1'然后tempdata等待3 *期;
elsif swt_a ='1'然后tempdata等到swt_a ='0';
elsif swt_b ='1'然后tempdata等到swt_b ='0';
elsif swt_c ='1'然后tempdata等到swt_c ='0';
结束如果;结束过程;

dataout 以下为原文

I have created a sound box on my FPGA using VHDL. It has 3 push buttons ans 3 selector switches. When any selector switch is high '1' the tone tied to that switch plays as long as it is high. When it goes low the tone stops.

I would like my buttons, when switched high '1', to stay high for 3 second after the button is released, then go to zero. Some call this a one-shot, where a single cycle latches a bit. I am trying to use a wait statement, but it is not working. The tone plays as long as I hold the button, but stops immediately when released.

Here is a section of my code. Any help is appreciated.

NOTE:  period is a constant  (i.e. 'constant period: time := 1000ms;')

D1: process
begin
if btn_a = '1' then
tempdata <= squareout;
wait for 3 * period;
elsif btn_b = '1' then
tempdata <= integerout;
wait for 3 * period;
elsif btn_c = '1' then
tempdata <= ipcoreout;
wait for 3 * period;
elsif swt_a = '1' then
tempdata <= squareout;
wait until swt_a = '0';
elsif swt_b = '1' then
tempdata <= integerout;
wait until swt_b = '0';
elsif swt_c = '1' then
tempdata <= ipcoreout;
wait until swt_c = '0';
end if;
end process;
dataout <= tempdata;